The smallest value of n that satisfies this inequality is n = 11. Therefore, we need at least 11 terms to obtain an approximation of the series IM8 12ke 0.49k2 accurate to 10-6.
To find the smallest number of terms needed to obtain an approximation of the series IM8 12ke 0.49k2 accurate to 10-6, we need to use the formula for the partial sum of a series. The partial sum of the given series up to n terms is:
S(n) = IM8 + 12e + 0.49(2^2) + 0.49(3^2) + ... + 0.49(n^2)
We want to find the smallest value of n such that the error between S(n) and the true value of the series is less than 10^-6. The error between S(n) and the true value of the series can be approximated by the absolute value of the next term in the series:
|an+1| = 0.49((n+1)^2)
So we need to find the smallest value of n such that:
|an+1| < 10^-6
0.49((n+1)^2) < 10^-6
(n+1)^2 < (10^-6)/0.49
n+1 < sqrt((10^-6)/0.49)
n < sqrt((10^-6)/0.49) - 1
n < 11.75
Since n must be a whole number, the smallest value of n that satisfies this inequality is n = 11. Therefore, we need at least 11 terms to obtain an approximation of the series IM8 12ke 0.49k2 accurate to 10-6.
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Pls help y’all I’m struggling
The area of the square is 49 in². (third option)
The area of the circle is 75.39 in². (fourth option)
The area of the shaded portion is 26.39 in².(first option)
What are the area of the shapes?A square is a quadrilateral with four equal sides.
Area of a square = length²
7² = 49 in²
A circle is a bounded figure which points from its center to its circumference is equidistant.
Area of a circle = πr²
Where :
π = pi = 3.14R = radius3.14 x 4.9² = 75.39 in²
Area of the shaded portion = area of circle - area of square
75.39 in² - 49 in² = 26.39 in²
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What would be an example of a tiered observation if you are measuring temperature?
Ranking the temperatures.
Tiered observations only apply to discrete variables.
Measurements rounded off to the nearest degree.
Only considering those temperatures in a certain tier.
An example of a tiered observation when measuring temperature would be measurements rounded off to the nearest degree.
This means that when you observe and measure the temperature, you would round the values to the nearest whole degree, providing a concise and uniform set of data points for analysis or comparison.
One example of a tiered observation when measuring temperature could be only considering those temperatures in a certain range, such as only observing temperatures that fall within the range of 60-70 degrees Fahrenheit. This would be a tiered observation because it is limiting the range of data being observed.
However, it's important to note that measurements rounded off to the nearest degree could also be considered a tiered observation because it's grouping data into discrete categories based on the rounding method used.
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which of the following situations can use the binomial probability distribution? group of answer choices a sampling of 100 parts to determine whether or not they meet specifications.
The situation that can use the binomial probability distribution is a sampling of 100 parts to determine whether or not they meet specifications.
The binomial probability distribution is used to model the probability of a certain number of successes in a fixed number of independent trials, where each trial has only two possible outcomes: success or failure. In the given situation, each part in the sample either meets the specifications (success) or does not (failure), which makes it a binomial experiment.
To use the binomial probability distribution, we need to know the probability of success (p) and the number of trials (n). In the given situation, we can determine the probability of a part meet specifications based on the given specifications, and the number of trials is fixed at 100, as we are sampling 100 parts.
Using the binomial probability distribution, we can calculate the probability of a certain number of parts meeting specifications out of the 100 sampled parts. This can be useful in determining whether the sample meets the expected specifications or if there are any issues with the manufacturing process.
In summary, the binomial probability distribution can be used in the given situation of sampling 100 parts to determine whether or not they meet specifications, as it involves a fixed number of independent trials with only two possible outcomes.
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Is (-2, 7) a solution to the equation y = -5x - 3?
Answer:
yes
Step-by-step explanation:
let x be -2 and y be 7.
Then,
y = -5x-3
7= -5*(-2)-3
7= 10-3
7=7.
look at the following input/output table and mapping. Determine if the relation is a function and why.
No, It is not a function because the x's (input) do repeat.
We know that;
A relation between a set of inputs having one output each is called a function.
Now, We have to given that;
In the given relation,
The x's (input) do repeat.
Hence, It is not a function.
Thus, Correct statement is,
No, It is not a function because the x's (input) do repeat.
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3.
x² + 2x = 1
A. List the values for a, b, and c from the quadratic above (hint: c is not 1!)
a=
b=
C =
B. Fill in the values of a, b, and c to the quadratic formula below
X=
-( ) ± √(
2(
)²-4(
)
)( )
C. Simplify each section (one number) of the quadratic formula from part B
(note that we have split the formula into two problems because of the ± symbol)
and x
Answer:
a) a = 1 ; b = 2 ;c =-1
c) -1 + √2 ; -1 - √2
Step-by-step explanation:
Solving a quadratic equation using quadratic formula:x² + 2x = 1
a) x² + 2x - 1 = 0
Compare with ax² + bx + c = 0
a = 1 ; b = 2 and c = -1
b)
[tex]\boxed{x=\dfrac{-b \± \sqrt{b^2-4ac}}{2a}}[/tex]
[tex]= \dfrac{-2 \± \sqrt{2^2-4*1*(-1)}}{2*1}\\\\\\\\C) \ =\dfrac{-2 \± \sqrt{4+4}}{2}\\\\=\dfrac{-2 \± \sqrt{8}}{2}\\\\=\dfrac{-2 \±2\sqrt{2}}{2}\\\\=\dfrac{2(-1 \± \sqrt{2})}{2}\\\\= -1 \± \sqrt{2}[/tex]
x = -1 + √2 or x = -1 -√2
Assignment Booklet 5 2 thematics 30-2 2. Solve each equation and identify the non-permissible values. Record the answers as exact values (no decimals!) harks) 2. a. 1 + X 4 х 1 b. 3n-1 3n +6 + 2 n
The non-permissible values are any values of n that would make the original equation undefined. In this case, there are no such non-permissible values.
a.
The given equation is:
[tex]1 + x^4 = x[/tex]
Rearranging terms, we get:
[tex]x^4 - x + 1 = 0[/tex]
To solve this equation, we can use the quartic formula:
x = [ -b ± sqrt( b^2 - 4ac ) ] / 2a
Here, a = 1, b = -1, and c = 1, so we have:
x = [ -(-1) ± sqrt( (-1)^2 - 4(1)(1) ) ] / 2(1)
x = [ 1 ± sqrt( -3 ) ] / 2
Since the discriminant is negative, the solutions are complex:
x = [ 1 ± i*sqrt(3) ] / 2
Therefore, the non-permissible values are any values of x that would make the original equation undefined. In this case, there are no such non-permissible values.
b.
The given equation is:
3^(n-1) / (3^n + 6) + 2^n = 0
To solve for n, we can start by simplifying the first term:
3^(n-1) / (3^n + 6) = 3^(-1) / (1 + 2*3^(-n))
Substituting this into the original equation, we get:
3^(-1) / (1 + 2*3^(-n)) + 2^n = 0
Multiplying both sides by (1 + 23^(-n)), we get:
3^(-1) + 2^n(1 + 2*3^(-n)) = 0
Simplifying, we get:
2^n + 2*3^(n-1) = 0
Dividing both sides by 23^(n-1), we get:
(1/2)(1/3)^n + 1 = 0
Multiplying both sides by -2 and taking the logarithm of both sides, we get:
n = log(2/3) / log(3) - log(2)
Therefore, the solution is:
n = log(2/3) / log(3) - log(2)
The non-permissible values are any values of n that would make the original equation undefined. In this case, there are no such non-permissible values.
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You measure 21 textbooks weights, and find they have a mean weight of 72 ounces. Assume the population standard deviation is 5.4 ounces. Based on this, construct a 90% confidence interval for the true
The 90% confidence interval for the true mean weight of the textbooks is approximately (70.062 ounces, 73.938 ounces).
Given that you measured 21 textbooks and found a mean weight of 72 ounces with a population standard deviation of 5.4 ounces, we can follow these steps:
1. Identify the sample size (n), sample mean (X), population standard deviation (σ), and confidence level (90%).
n = 21
X = 72 ounces
σ = 5.4 ounces
Confidence level = 90%
2. Determine the critical value (z) for a 90% confidence interval. For a 90% confidence interval, the critical value (z) is 1.645.
3. Calculate the standard error (SE) using the formula [tex]SE = \frac {σ }{\sqrt{n} }[/tex].
[tex]SE = \frac{5.4}{\sqrt{21} } = 1.177[/tex]
4. Calculate the margin of error (ME) using the formula ME = z * SE.
ME = 1.645 * 1.177 = 1.938
5. Construct the confidence interval using the formula: X ± ME.
Lower limit = 72 - 1.938 = 70.062
Upper limit = 72 + 1.938 = 73.938
Based on your measurements, the 90% confidence interval for the true mean weight of the textbooks is approximately (70.062 ounces, 73.938 ounces).
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Find mLMN.
5 cm
N
M
L
14.3 cm
Applying the formula for the length of an arc, the measure of angle LMN is approximately: 164°.
What is the Length of an Arc?The length of an arc (s) = ∅/360 × 2πr, where r is the radius of the circle.
Given the following from the image attached below, we have:
Reference angle (∅) = m<LMN
length of an arc (s) = 14.3 cm
Radius (r) = 5 cm
Plug in the values:
∅/360 × 2π × 5 = 14.3
∅/360 × 10π = 14.3
∅/360 = 14.3/10π
∅ = 14.3/10π × 360
∅ ≈ 164°
The measure of angle LMN ≈ 164°
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A traffic engineer developed the continuous function R, graphed above, to model the rate at which vehicles pass a certain intersection over an 8-hour time period, where R(t) is measured in vehicles per hour and t is the number of hours after 6:00 AM. According to the model, how many vehicles pass the intersection between time t = 0 and time t = 8? A. 1400 B. 1600 C. 14,400 D. 44,800
the total area under the curve is 2400.
To find the number of vehicles that pass the intersection between time t = 0 and time t = 8, we need to calculate the definite integral of the function R(t) from t = 0 to t = 8:
∫(0 to 8) R(t) dt
Looking at the graph of R(t), we can see that it consists of two parts: a rectangle with base 2 and height 600, and a triangle with base 6 and height 400. The area of the rectangle is 2 x 600 = 1200, and the area of the triangle is (1/2) x 6 x 400 = 1200. Therefore, the total area under the curve is 2400.
So, the number of vehicles that pass the intersection between time t = 0 and time t = 8 is:
∫(0 to 8) R(t) dt = 2400
Since R(t) is measured in vehicles per hour, this means that 2400 vehicles pass the intersection between time t = 0 and time t = 8. Therefore, the answer is 2400, which is not one of the given answer choices.
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y1= sin(x) y2= sin(2x) y3= sin(3x) % 1st plot is to be red and dashed % 2nd plot is to be blue and solid % 3rd plot is to be black and dotted The plot ranges from -6 to 6 in steps of 0.2. Use enough steps so that the plots are smooth. You must include a title ('Multiple Plots'), X-axis label ('x'), y-axis label ('Sine functions') and also a legend ('sin(x)', 'sin(2x)', 'sin(3x)'). In additional, use the grid on and axis equal command.
To create a plot with three different sine functions, we can use MATLAB code that includes the "plot" function, as well as specific parameters to set the ranges, colors, and styles of each line. First, we need to set up the x-axis range using the "range" function, which takes in the minimum, maximum, and step size values. In this case, we want the range to be from -6 to 6 with steps of 0.2, so we can write:
x = (-6:0.2:6);
Next, we can define each y value using the sine function and the corresponding multiple of x. For example, y1 corresponds to sin(x), so we can write:
y1 = sin(x);
Similarly, we can define y2 and y3 as:
y2 = sin(2*x);
y3 = sin(3*x);
Now, we can use the "plot" function to create a graph with all three sine functions plotted together. We want each function to be plotted with a different color, style, and legend label, so we can specify these parameters in the "plot" function call. Specifically, we want:
- y1 to be plotted in red and dashed
- y2 to be plotted in blue and solid
- y3 to be plotted in black and dotted
- a legend to be added with the labels 'sin(x)', 'sin(2x)', and 'sin(3x)'
- a title to be added with the label 'Multiple Plots'
- an x-axis label to be added with the label 'x'
- a y-axis label to be added with the label 'Sine functions'
Here is the complete code:
x = (-6:0.2:6);
y1 = sin(x);
y2 = sin(2*x);
y3 = sin(3*x);
plot(x, y1, 'r--', 'LineWidth', 1.5, 'DisplayName', 'sin(x)');
hold on;
plot(x, y2, 'b-', 'LineWidth', 1.5, 'DisplayName', 'sin(2x)');
plot(x, y3, 'k:', 'LineWidth', 1.5, 'DisplayName', 'sin(3x)');
title('Multiple Plots');
xlabel('x');
ylabel('Sine functions');
legend('show', 'Location', 'northwest');
grid on;
axis equal;
The "hold on" command ensures that all three plots are shown on the same graph. The "LineWidth" parameter sets the width of each line, and the "DisplayName" parameter sets the label for each line in the legend. Finally, the "grid on" and "axis equal" commands add a grid to the graph and ensure that the x and y axes are scaled equally.
Overall, this code will create a graph with three smooth sine functions plotted together, each with a different color, style, and legend label.
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The effectiveness of a blood-pressure drug is being investigated. An experimenter finds that, on average, the reduction in systolic blood pressure is 53.9 for a sample of size 24 and standard deviation 5.6. Estimate how much the drug will lower a typical patient's systolic blood pressure (using a 90% confidence level). Assume the data is from a normally distributed population. Enter your answer as a tri-linear inequality accurate to three decimal places.
_______ < μ < _________ please teach using calculator method
We can be 90% confident that the true mean reduction in systolic blood pressure for the population is between 51.433 and 56.367.
To estimate the confidence interval for the true mean reduction in systolic blood pressure, we can use the formula:
CI = X ± Zα/2 * (σ/√n)
where X is the sample mean, Zα/2 is the critical value of the standard normal distribution for a given level of confidence (α), σ is the population standard deviation, and n is the sample size.
In this case, X = 53.9, σ = 5.6, n = 24, and we want a 90% confidence interval, so α = 0.1 and Zα/2 = 1.645 (using a standard normal distribution table or calculator).
Substituting the values, we get:
CI = 53.9 ± 1.645 * (5.6/√24)
CI = 53.9 ± 2.467
CI = (51.433, 56.367)
Therefore, we can be 90% confident that the true mean reduction in systolic blood pressure for the population is between 51.433 and 56.367.
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determine the minimum area of a poster that will contain 50 square inches of printed material and have 4 inch margins on the top and bottom and 2 inch margins on the left and right.
The minimum area of a poster that will contain 50 square inches of printed material and have 4-inch margins on the top and bottom and 2-inch margins on the left and right is 98 square inches.
To determine the minimum area of the poster, we need to consider the dimensions of the printed material and the margins. The printed material covers an area of 50 square inches. The margins on the top and bottom are each 4 inches, which means we need to add 8 inches to the height of the printed material. The margins on the left and right are each 2 inches, which means we need to add 4 inches to the width of the printed material.
To determine the minimum area of the poster, follow these steps:
1. Add the top and bottom margins to the height of the printed material:
Height = Printed Height + Top Margin + Bottom Margin
Height = 50 square inches (assuming printed material height is 1 inch) + 4 inches + 4 inches
Height = 9 inches
2. Add the left and right margins to the width of the printed material:
Width = Printed Width + Left Margin + Right Margin
Width = 50 square inches / 1 inch (since printed material height is 1 inch) + 2 inches + 2 inches
Width = 54 inches
So, the total height of the poster is the height of the printed material plus the top and bottom margins, which is 50/width + 8 inches. The total width of the poster is the width of the printed material plus the left and right margins, which is 50/height + 4 inches.
Therefore, the minimum area of the poster is 7 x 14 = 98 square inches.
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5 1/3 divided by 3/4
Answer:
The answer to your problem is, [tex]7\frac{1}{9}[/tex]
Step-by-step explanation:
Calculation process:
= [tex]\frac{16}{3}[/tex] ÷ [tex]\frac{3}{4}[/tex]
= [tex]\frac{16}{3}[/tex] × [tex]\frac{4}{3}[/tex]
= [tex]\frac{16*4}{3*3}[/tex]
= [tex]\frac{64}{9}[/tex] = [tex]7\frac{1}{9}[/tex]
Thus the answer to your problem is, [tex]7\frac{1}{9}[/tex]
Solve the quadratic equation
7. 3x2 + 13x10 = 0
9. 12n²-11n +2=0
11. 4x² + 12x +9=0
X
8. 5x28x +3=0
10. 10a²a-2=0
12. 8x2 10x + 3 = 0
The solution of the quadratic equations are shown below.
How do you solve the quadratic equation?There are various methods that we could use when we want to solve a quadratic equation and these include;
1) Formula method
2) Graphical method
3) Completing the square method
4) Factor method
We have solved the following quadratic equations by factoring.
1) 3x^2 + 13x +10 = 0
x = - 1 and -10/3
2) 12n²-11n +2=0
n = 2/3 and 1/4
3) 5x^2 + 8x +3=0
x = -3/5 and -1
4) 10a²+ a -2=0
a = 2/5 and -1/2
5) 8x^2 + 10x + 3 = 0
x = -1/2 and -3/4
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Evaluate the Expression
You want to hang 6 pictures in a row on a wall. You have 11 pictures from which to choose. How many picture arrangements are possible?
The number of different arrangements that can be formed is 7920
How many different arrangements can be formed?From the question, we have the following parameters that can be used in our computation:
Pictures = 11
Arranged pictures = 6
These can be represented as
n = 11 and r = 6
The number of different arrangements that can be formed is
Number = nPr
So, we have
Number = 11P4
Evaluate
Number = 7920
Hence, the arrangements are 7920
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Find the mean, median, interquartile range and mean absolute deviation of the set of numbers. Round to the nearest tenth, if necessary. 1, 1, 4, 8, 9, 3, 8 please help
Answer:
mean- 4.9
median- 4
interquartile range- 7
Step-by-step explanation:
Hope this helps! :)
Use the equation −20x+3x 2−7=0 to answer all of the following questions.
5. At practice, a soccer athlete warmed up for 8 minutes, participated in drills for 45
minutes, and scrimmaged for 17 minutes. How many total minutes of activity did this
soccer athlete complete during this practice?
110
Answer:
The soccer athlete completed 70 minutes of activity during this practice.
Step-by-step explanation:
mark brainliest
philosophy question. introduction of logic. please answer correctlyand fully.Use the first thirteen rules of inference to derive the conclusions of the following symbolized arguments: A. 1. X-M 2. M |-(v-M) B. 1. L» (B v 0) 2.-(-0.-B) S Los C. 1. Ko (Y.Z) 2. (K.C) v ( MK) Z
Using the first thirteen rules of inference to derive the conclusions of the given symbolized arguments.
A. Conclusion: v-X
X-M (Premise)M (Assumption for Conditional Proof)|-(v-M) (Premise)v (Disjunctive Syllogism: 3, M)v-X (Conditional Proof: 2-4)B. Conclusion: L
L» (B v 0) (Premise)-(-0.-B) (Premise)-0 v -(-B) (Material Implication: 2)-0 v B (Double Negation: 3)B v 0 (Commutation: 1)-B»0 (Material Implication: 5)-B (Disjunctive Syllogism: 4,6)0 (Modus Ponens: 6,7)L (Disjunctive Syllogism: 1,8)C. Conclusion: Z
Ko (Y.Z) (Premise)(K.C) v ( MK) (Premise)-Z (Assumption for Conditional Proof)-(Y.Z) (Material Implication: 1)-Y v -Z (De Morgan's Law: 4)Y (Disjunctive Syllogism: 2, MK)-Z v -Z (Addition: 3)Z (Negation Elimination: 7)In the first argument, the disjunctive syllogism and conditional proof rules were used to derive the conclusion. In the second argument, the material implication, double negation, commutation, modus ponens, and disjunctive syllogism rules were used to derive the conclusion. In the third argument, the material implication, De Morgan's Law, disjunctive syllogism, addition, and negation elimination rules were used to derive the conclusion.
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Using the rules of inference to derive the conclusions of the given symbolized arguments.
A. Conclusion: v-X
X-M (Premise)
M (Assumption for Conditional Proof)
|-(v-M) (Premise)
v (Disjunctive Syllogism: 3, M)
v-X (Conditional Proof: 2-4)
B. Conclusion: L
L» (B v 0) (Premise)
-(-0.-B) (Premise)
-0 v -(-B) (Material Implication: 2)
-0 v B (Double Negation: 3)
B v 0 (Commutation: 1)
-B»0 (Material Implication: 5)
-B (Disjunctive Syllogism: 4,6)
0 (Modus Ponens: 6,7)
L (Disjunctive Syllogism: 1,8)
C. Conclusion: Z
Ko (Y.Z) (Premise)
(K.C) v ( MK) (Premise)
-Z (Assumption for Conditional Proof)
-(Y.Z) (Material Implication: 1)
-Y v -Z (De Morgan's Law: 4)
Y (Disjunctive Syllogism: 2, MK)
-Z v -Z (Addition: 3)
Z (Negation Elimination: 7)
How to explain the informationThe disjunctive syllogism and conditional proof procedures were employed to reach the conclusion in the first argument.
The material implication, double negation, commutation, modus ponens, and disjunctive syllogism rules were employed to reach the conclusion in the second argument. The material implication, De Morgan's Law, disjunctive syllogism, addition, and negation elimination rules were utilized to obtain the conclusion in the third argument.
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according to an avid aquarist, the average number of fish in a 20-gallon tank is 10, with a standard deviation of two. his friend, also an aquarist, does not believe that the standard deviation is two. she counts the number of fish in 15 other 20-gallon tanks. based on the results that follow, do you think that the standard deviation is different from two at the 5% level? data: 11; 10; 8; 10; 10; 11; 11; 10; 12; 9; 8; 9; 11; 10; 11.
We have evidence to suggest that the standard deviation of the number of fish in a 20-gallon tank is different from twoat the 5% level.
To determine if the standard deviation is different from two at the 5% level, we can perform a hypothesis test. The null hypothesis is that the standard deviation is two, and the alternative hypothesis is that the standard deviation is different from two.
We can use a chi-square test statistic to test this hypothesis. The test statistic is calculated as:
χ² = (n - 1) * s² / σ²
where n is the sample size, s is the sample standard deviation, and σ is the hypothesized population standard deviation.
We can then compare this test statistic to the critical value from the chi-square distribution with n - 1 degrees of freedom at the 5% significance level.
Using the given data, we have:
n = 15
s = 1.256
σ = 2
Plugging these values into the formula, we get:
χ² = (15 - 1) * 1.256² / 2² = 42.891
Using a chi-square distribution table or a calculator, we find that the critical value with 14 degrees of freedom at the 5% level is 23.685.
Since our test statistic (42.891) is greater than the critical value (23.685), we reject the null hypothesis and conclude that the standard deviation is different from two at the 5% level.
Therefore, based on the data provided, we have evidence to suggest that the standard deviation of the number of fish in a 20-gallon tank is different from two.
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!!!!!! I need help asap
Answer:
3.2
Step-by-step explanation:
Answer:
pythagorean theorem !!!!!!
Step-by-step explanation:
0.8²+1.6²=3.2²
0.64+2.56=10.24
THAT means the distance is
[tex] \sqrt{10.24} [/tex]
=3.2miles
CARD 4
Zoe opens a savings account that
earns annual compound interest. If she
doesn't make any deposits or
withdrawals after her initial deposit,
the balance in the account after x
years can be represented by the
equation below.
b(x)=675(1.045)*
D
Duncan says the
balance in the
account increases at
a rate of 45% each
year.
Daniella says the
balance in the
account increases at
a rate of 4.5% each
year.
Which answer is right
The correct answer about the savings account that Zoe opened, represented by the equation b(x)=675(1.045)ˣ is B. Daniella says the balance in the account increases at a rate of 4.5% each year.
What is an equation?An equation is a mathematical statement that two or more algebraic expressions (the combination of variables with constants and mathematical operands) are equal or equivalent.
This equation is known as the future value equation, formula, or function.
Initial investment = $675
Compound interest rate = 4.5% (0.045 x 100)
Future value factor = 1.045 (100% + 4.5%)
Based on the future value function above, we can conclude that the compound interest rate is 4.5%, which is equivalent to 0.045 as given in the equation.
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how to rationalise root 3-1/5
[tex] \sqrt{ \frac{3 - 1}{5} } = \sqrt{ \frac{2}{5} } = \frac{ \sqrt{2} }{ \sqrt{5} } = \frac{ (\sqrt{2} )}{ (\sqrt{5}) } \frac{( \sqrt{5})}{ (\sqrt{5} )} = \frac{ \sqrt{10} }{5} [/tex]
The answer is V10/5
-4/7-8/9+4/7+9/8
Which of the following expressions are equivalent to
The equivalent expression for -4/7-8/9+4/7+9/8 is given by option C. -8/9 + 9/8 and option D. - (4/7+ 8/9 ) + 4/7 + 9/8.
The expression is equals to,
-4/7-8/9+4/7+9/8
Verify all attached options using property of addition.
-4/7- (8/9+4/7 )+9/8
Open the parenthesis as plus minus is minus we get,
- 4/7 - 8/9 - 4 /7 + 9/8
It is not equivalent to -4/7-8/9+4/7+9/8.
Incorrect option.
- ( 4/7-8/9+4/7 ) + 9/8
Open the parenthesis as (+)( - )is minus and ( - ) ( - ) is plus we get,
- 4/7 + 8/9 - 4 /7 + 9/8
It is not equivalent to -4/7-8/9+4/7+9/8.
Incorrect option.
-8/9 + 9/8
= 0 -8/9 + 9/8
= -4/7 + 4/7 -8/9 + 9/8
Rearrange terms we get,
-4/7-8/9+4/7+9/8
It is equivalent to -4/7-8/9+4/7+9/8
Correct option.
- (4/7+ 8/9 ) + 4/7 + 9/8
Open the parenthesis as plus minus is minus we get,
-4/7 -8/9 + 4/7 + 9/8
It is equivalent to -4/7-8/9+4/7+9/8
Correct option.
0
Incorrect option.
Therefore, for the given expression equivalent terms are option C. -8/9 + 9/8 and option D. - (4/7+ 8/9 ) + 4/7 + 9/8.
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The above question is incomplete, the complete question is:
-4/7-8/9+4/7+9/8
Which of the following expressions are equivalent to
Attached options.
A researcher has done a study to look at wether senior citizens sleep fewer hours than the general population. She has gathered data on 30 senior citizens regarding how many hours of sleep they get each night. She performs a two-tailed single-sample t test with a .05 alpha level on her results. She calculates her obtained statistic (tobt) = -1.98. Tcrit for a two tailed t test with an alpha level of .05 and with df=29 is +/-2.045. What decision should she make? a. Fail to Reject/Retain the null. absolute value of tobt > absolute value of tcrit b. Reject the null absolute value of tobt> absolute value of tcrit c. Fail to Reject/Retain the null. absolute value of tobt
Based on the information provided, the researcher should choose option a, which is to fail to reject/retain the null hypothesis. This is because the absolute value of the obtained statistic (tobt) (-1.98) is less than the absolute value of the critical value (tcrit) for a two-tailed t test with an alpha level of .05 and with df=29 (which is +/-2.045).
To clarify some of the terms used, the researcher in this scenario is conducting a hypothesis test to compare the population of senior citizens' average hours of sleep to that of the general population. She collected a sample of 30 senior citizens to represent the population. The null hypothesis is the statement that there is no difference between the two populations in terms of average hours of sleep. The alternative hypothesis is the statement that the senior citizens sleep fewer hours than the general population. The obtained statistic (tobt) is a measure of how far the sample mean deviates from the null hypothesis. The critical value (tcrit) is the cutoff value used to determine whether the obtained statistic is significant enough to reject the null hypothesis.
c. Fail to Reject/Retain the null. absolute value of tobt < absolute value of tcrit
Explanation:
The researcher performed a two-tailed single-sample t-test to compare the sleep hours of a sample of 30 senior citizens with the general population. The obtained statistic (tobt) is -1.98, and the critical value (Tcrit) for this test with an alpha level of .05 and df=29 is +/-2.045.
To make a decision, we compare the absolute values of tobt and tcrit:
Absolute value of tobt: |-1.98| = 1.98
Absolute value of tcrit: 2.045
Since the absolute value of tobt (1.98) is less than the absolute value of tcrit (2.045), we fail to reject the null hypothesis. This means the researcher cannot conclude that there is a significant difference in sleep hours between senior citizens and the general population based on her sample.
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Use the information given in Exercises 8 - 10 to find the necessary confidence bound for the binomial proportion P. Interpret the interval that you have constructed. 99% upper bound, n = 55, x = 24
The 99% upper bound for the binomial proportion P is 0.790. To find the necessary confidence bound for the binomial proportion P, we can use the formula: Upper bound = x/n + Zα/2√(x/n(1-x/n))
In this case, we are looking for a 99% upper bound, so Zα/2 = 2.576. Plugging in the given values, we get:
Upper bound = 24/55 + 2.576√(24/55(1-24/55))
= 0.526 + 2.576(0.100)
= 0.790
Therefore, the 99% upper bound for the binomial proportion P is 0.790.
Interpreting the interval, we can say that we are 99% confident that the true proportion of whatever we are measuring (which is represented by P) is no higher than 0.790. In other words, we can be fairly certain that the actual proportion falls within the interval from 0 to 0.790.
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A cone with radius 6 feet and height 15 feet is shown.
6
ft
Enter the volume, in cubic feet, of the cone. Round your
answer to the nearest hundredth. step by step expinayion and answer with check
Answer:
The volume of a cone is given by the formula:
V = (1/3)πr^2h
where r is the radius of the base, h is the height, and π is the constant pi (approximately 3.14).
Plugging in r = 6 and h = 15, we get:
V = (1/3)π(6^2)(15) = 540π cubic feet
Rounding to the nearest hundredth, we get:
V ≈ 1696.63 cubic feet
Therefore, the volume of the cone is approximately 1696.63 cubic feet.
To check, we can use the formula for the volume of a cone to calculate the volume using different methods. For example, we can use the fact that the cone is one-third the volume of a cylinder with the same base and height. The cylinder has radius 6 feet and height 15 feet, so its volume is:
V_cylinder = π(6^2)(15) = 540π cubic feet
Dividing by 3, we get:
V_cone = (1/3)V_cylinder = (1/3)(540π) = 180π cubic feet
Rounding this to the nearest hundredth, we get:
V_cone ≈ 565.49 cubic feet
This is reasonably close to our previous answer of 1696.63 cubic feet, so we can be confident that our calculation is correct.
Step-by-step explanation:
Why are t value larger than the corresponding z value?
The t-value is generally larger than the corresponding z-value because the t-distribution has heavier tails than the standard normal distribution. This means that the t-distribution has more probability in the tails than the standard normal distribution. As a result, the critical values for the t-distribution are larger than the corresponding critical values for the standard normal distribution.
Another reason for this difference is that the t-distribution takes into account the variability of the sample mean, which is estimated using the sample standard deviation. In contrast, the standard normal distribution assumes that the population standard deviation is known and fixed.
When the sample size is small, the t-distribution is more appropriate because it accounts for the additional uncertainty introduced by estimating the population standard deviation from the sample. As the sample size increases, the t-distribution approaches the standard normal distribution, and the difference between the t-value and the corresponding z-value decreases.
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Howto prove for root test convergence for complex number.
To prove convergence for the root test with complex numbers, we use the same approach as with real numbers.
Let's consider a series ∑an with complex terms. We can apply the root test by taking the nth root of the absolute value of each term, which gives us:
lim (n→∞) ∛|an|
If this limit is less than 1, then the series converges absolutely. If it is greater than 1, then the series diverges.
To prove convergence for the root test, we need to show that this limit is less than 1. We can do this by expressing the complex number an in polar form, such that an = rn*e^(iθn), where rn is the magnitude of an and θn is its argument.
Then, taking the nth root of the absolute value of an, we get:
|an|^1/n = (rn)^(1/n)
We can express rn as |an|*cos(θn) + i*|an|*sin(θn), and take the nth root of each term separately:
|an|^1/n = [(|an|*cos(θn))^2 + (|an|*sin(θn))^2]^(1/2n)
= |an|^(1/n) * [(cos(θn))^2 + (sin(θn))^2]^(1/2n)
= |an|^(1/n)
Since the limit of |an|^(1/n) is the nth root of the magnitude of the series, we can rewrite the root test as:
lim (n→∞) ∛|an| = lim (n→∞) |an|^(1/n)
If we can show that this limit is less than 1, then we have proven convergence for the root test with complex numbers.
One way to do this is to use the fact that |an|^(1/n) ≤ r, where r is the radius of convergence of the series. This inequality follows from Cauchy's root test, which applies to both real and complex numbers.
Therefore, if the radius of convergence of the series is less than 1, then the limit of |an|^(1/n) is also less than 1, and the series converges absolutely.
In summary, to prove convergence for the root test with complex numbers, we express each term in polar form and take the nth root of its magnitude. We then show that the limit of these roots is less than 1 by using Cauchy's root test and the radius of convergence of the series.
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