1) Given that the "hot hand theory" states that after success comes right away a series of successful instances further and further.
We can state that this is a fallacy, an erroneous logical argument since the future success is not a dependent event from the current or previous successful instances.
And also, success not a predictable event given that this notion of success is so broad and vague and peculiar and envolves so many other variables and aspects that we can not state.
Find the area of the region bounded by the courses y = sin^-1(x/6), y = 0, and x = 6 obtained by integrating with respect to y. Your work must include the definite integral and the antiderivative.
The equation is given
[tex]y=sin^{-1}(\frac{x}{6})[/tex]y=0 and x=6.
RequiredTo determine the area of region bounded by courses of the equation given.
Explanation[tex]\begin{gathered} y=sin^{-1}(\frac{x}{6})\Rightarrow\Rightarrow siny=\frac{x}{6} \\ 6siny=x \end{gathered}[/tex][tex]x=6,y=0[/tex]Therefore
[tex]\begin{gathered} 6siny=6 \\ siny=1 \\ y=\frac{\pi}{2} \end{gathered}[/tex][tex]\begin{gathered} 6siny=0 \\ siny=0 \\ y=0,\pi \end{gathered}[/tex]Now integrate the equation.
[tex]\int_0^{\frac{\pi}{2}}(6-6siny)dy=(6y+6cosy)_0^{\frac{\pi}{2}}[/tex][tex](\frac{6\pi}{2}+6cos\frac{\pi}{2})-(6\times0+6cos0)=3\pi-6[/tex]AnswerHence the area of region bounded by the courses is
[tex]3\pi-6[/tex]Answer this question and show me how to check it
Given:
[tex]\text{length =}8\times10^4\text{ m}[/tex][tex]t\text{hickness}=5\times10^{-6}m[/tex]To find the standard form of length, multiply and divide the length by 10, we get
[tex]\text{length =}8\times10^4\text{ }\times\frac{10}{10}[/tex][tex]\text{length =}\frac{8}{10}\times10^4\text{ }\times10[/tex][tex]\text{Use }\frac{8}{10}=0.8\text{ and }10^4\times10=10^{4+1}=10^5.[/tex][tex]\text{length =}0.8\times10^5m[/tex]To find the standard form of thickness, multiply and divide the thickness by 10, we get
[tex]t\text{hickness}=5\times10^{-6}\times\frac{10}{10}[/tex][tex]t\text{hickness}=\frac{5}{10}\times10^{-6}\times10[/tex][tex]\text{Use }\frac{5}{10}=0.5\text{ and }10^{-6}\times10=10^{-6+1}=10^{-5}.[/tex][tex]t\text{hickness}=0.5\times10^{-5}m[/tex]Hence the standard form of length and thickness is
[tex]\text{length =}0.8\times10^5m[/tex][tex]t\text{hickness}=0.5\times10^{-5}m[/tex]A system of equations is created by using the line represented by 2x+4y= 0 and the line represented by the data in the table below. x -1 3 5 6 у 8 -1 3 5 -10 6 -13 What is the x-value of the solution to the system?
To find what we are looking for we first need to find the second equation.
To do this we need to use the equation of a line:
[tex]y-y_1=m(x-x_1)[/tex]where (x1,y1) is a point on the line and m is the slope. The slope of a line is given by:
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]Using the first two points on the table we get:
[tex]\begin{gathered} m=\frac{-4-8}{3-(-1)} \\ =\frac{-12}{4} \\ =-3 \end{gathered}[/tex]Now that we have the slope we plug it in the equation of a line with the values of any of the points in the table (we are going to use the first one). Then:
[tex]\begin{gathered} y-8=-3(x-(-1)) \\ y-8=-3(x+1) \\ y-8=-3x-3 \\ 3x+y=5 \end{gathered}[/tex]Now that we have the equation of the second line we conclude that we have the system of equations:
[tex]\begin{gathered} 2x+4y=0 \\ 3x+y=5 \end{gathered}[/tex]To find the x value of the solution we solve the second equation for y, then:
[tex]y=5-3x[/tex]now we plug this value into the first equation and solve for x:
[tex]\begin{gathered} 2x+4(5-3x)=0 \\ 2x+20-12x=0 \\ -10x=-20 \\ x=\frac{-20}{-10} \\ x=2 \end{gathered}[/tex]Therefore, the x value of the solution of the system is 2.
Find the value of X in the image below. Enter your answer without any labels. fc B 399 (3x)º A Z D
We are asked to find the value of x in the following image:
Where we see that the addition of the angle that measures 39 degrees plus the angle indicated as 3x degrees, must give us a right angle (that is 90 degrees) in order to be the suplementary angle to a right angle formed at the intersection Z
So we can write the following equation:
39 + 3 x = 90
and solve for x by subtracting 39 from both sides
3 x = 90 - 39
3 x = 51
divide both sides by 3 in order to isolate x completely
x = 51 / 3
x = 17
a spinner has three unequal sections: red, yellow, and blue. the table shows the results of nolan’s spins (red:10, yellow:14, blue:6). find the experimental probability of landing on yellowanswer choices are 1/4 7/15 7/30 1/2
A spinner has three unequal sections.
Red = 10
Yellow = 14
Blue = 6
[tex]Total\: sections=Red+Yellow+Blue=10+14+6=30[/tex]We are asked to find the experimental probability of landing on yellow.
Recall that the experimental probability is given by
[tex]P=\frac{\text{number of desired outcomes}}{\text{total number of possible outcomes}}[/tex]In this case, the number of desired outcomes are
Yellow = 14
The total number of possible outcomes is
Total sections = 30
[tex]P=\frac{14}{30}=\frac{7}{15}[/tex]Therefore, the experimental probability of landing on yellow is 7/15
4x-6= 10x-3. solve for x
Write e^(½) = 1.6487 . . . in logarithm form.
SOLUTION
Given the question in the image, the following are the solution steps to answer the question.
STEP 1: Write the given exponential form
[tex]e^{\frac{1}{2}}=1.6487[/tex]STEP 2: Convert to logarithmic form
[tex]\begin{gathered} \operatorname{Re}call\text{ that, }e^x=y\Rightarrow\ln y=x \\ \text{Therefore, }e^{\frac{1}{2}}=1.6487\Rightarrow\ln 1.6487=\frac{1}{2} \end{gathered}[/tex]Hence, the logarithm form of the given exponential is written as:
[tex]\frac{1}{2}=\ln 1.6487[/tex]OPTION B
Jose bought a 20-pound bag of food for his dog. He fed his dog one-half of a pound of dog food each day. Which equation is used to determine, y, the amount of dog food that remains at the end of each day, x? Ay= 20 – 0.52 B.y= 20 + 0.51 cy= 200 – 0.5 Dy=202 + 0.5 Copyrig y
ok
y = amount of food dog that remains
x days
y = 20 - 0.5x This is the answer, It is letter
2.- slope = ((-100 - 160) / (4 - 8)
= -260 / -4
= 65 the answer is the third option
1. ) 14, 11, 8, 5, ... a. Identify each sequence as arithmetic or geometric, explain your answer. b. Find the next five terms. c. Write an explicit formula for the sequence. d. Write a recursive formula for the sequence. 2.) 2, 10, 50, 250, ... a. Identify each sequence as arithmetic or geometric, explain your answer. b. Find the next five terms. c. Write an explicit formula for the sequence. d. Write a recursive formula for the sequence.
The terms of the sequence are 14, 11, 8, 5, .....
Since
11 - 14 = -3
8 - 11 = -3
5 - 8 = -3
there is a constant difference between each two consecutive terms
then the sequence is an arithmetic sequence
The next five terms are
5 + -3 = 2
2 + -3 = -1
-1 + -3 = -4
-4 + -3 = -7
-7 + -3 = -10
The next five terms are
2, -1, -4, -7, -10
The explicit formula is
[tex]an=a+(n-1)d[/tex]an is any term in the sequence
a is the first term
d is the constant difference
n is the position of the number
a = 14
d = -3
[tex]an=14+(n-1)(-3)[/tex]Multiply -3 by the bracket
an = 14 + (-3)(n) - (-3)(1)
an = 14 - 3n + 3
Add the like terms
an = 17 - 3n
The explicit formula is an = 17 - 3n
The recursive formula is
First-term = 14, an = an-1 + d
First term = 14; an = an-1 + (-3)
[tex]a_1=14;a_n=a_{n-1}+(-3)[/tex]Given the following hypotheses:H0: u = 590H1: u ≠ 590A random sample of 15 observations is selected from a normal population. The sample mean was 595 and the sample standarddeviation 8. Using the 0.05 significance level:a. State the decision rule. Reject H0 when the teststatistic is outside the interval ( , )
Here, we are give the following:
Sample mean, x' = 595
Sample size, n = 15
Significance level = 0.05
Null and alternative hyporthesis:
H0: u = 590
H1: u ≠ 590
a) The decision rule:
The decision rule here, will be to reject the null hypothesis, H0, when the test statistic is outside the interval
Here, degrees of freedom, df = 15 -1 = 14
Therefore,
The critical values for t will be:
[tex] ( t_0._0_2_5_/_2, _1_4, t_1 -_0._0_2_5_/_2, _1_4,) [/tex]
Using the crtical value value table, we have:
at significance level of 0.025 and df = 14, t = -2.1448
at = 1 - 0.025 = 0.975, and df = 14, t = 2.1448
We can now say that:
The decision rule here, will be to reject the null hypothesis, H0, when the test statistic is outside the interval(-2.1448, 2.1448)
This means we are to reject H0 if t is greater than 2.1448 or when t is less than -2.1448
Determine whether there is a minimum or maximum value to the quadratic function.h(t) = −8t2 + 2t − 1Find the minimum or maximum value of h.Find the axis of symmetry.
For a quadratic function in the form:
[tex]f(x)=ax^2+bx+c[/tex]If a>0 the function opens up, it has a minimum
If a<0 the function opens down, it has a maximum
Axis of symmetry is x= -b/2a
Vertex: (-b/2a, f(-b/2a)), f(-b/2a) is the maximum or minimum value
For the given function:
[tex]h(t)=-8t^2+2t-1[/tex]a= -8
Parabola opens down. It has a maximum value
Find the axis of symmetry:
[tex]t=-\frac{2}{2(-8)}=\frac{-2}{-16}=\frac{1}{8}[/tex]Find the y-coordinate of the vertex:
[tex]\begin{gathered} h(\frac{1}{8})=-8(\frac{1}{8})\placeholder{⬚}^2+2(\frac{1}{8})-1 \\ \\ h(\frac{1}{8})=-8(\frac{1}{64})+\frac{2}{8}-1 \\ \\ h(\frac{1}{8})=-\frac{1}{8}+\frac{2}{8}-1 \\ \\ h(\frac{1}{8})=\frac{1}{8}-1 \\ \\ h(\frac{1}{8})=\frac{1-8}{8} \\ \\ h(\frac{1}{8})=-\frac{7}{8} \end{gathered}[/tex]______________
Then, the given function has a maximum value, the maximum value is -7/8, and the axis of symmetry is t=1/8We are standing on the top of a 384 feet tall building and launch a small object upward. The object's vertical position, measured in feet, after t seconds is h(t) = -16t^2 + 160t + 384. What is the highest point that the object reaches?
In linear equation, the highest point that the object reaches is x + y ≤ 20.
What in mathematics is a linear equation?
A linear equation is a first-order (linear) term plus a constant in the algebraic form y=mx+b, where m is the slope and b is the y-intercept.Sometimes, the aforementioned is referred to as a "linear equation of two variables," where x and y are the variables.The three main types of linear equations are the slope-intercept form, standard form, and point-slope form. In this post, we examine all three.h(t)=−16t2+64t+80
highest point
x co - ordinate = -96/2(-10) = 3 seconds
h ( 3) = 16(3)² + 76(3) + 112
= 256 ft
Learn more about linear equation
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#6 Solve each with quadratic formula simplify in radical form if needed
Given:
[tex]11v^2+8v=4[/tex]Sol:.
[tex]\begin{gathered} 11v^2+8v=4 \\ 11v^2+8v-4=0 \end{gathered}[/tex][tex]\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ x=\frac{-8\pm\sqrt[]{64-4(11)(-4)}}{2\times11} \\ x=\frac{-8\pm\sqrt[]{64+176}}{2\times11} \\ x=\frac{-8\pm15.49}{22} \\ x=\frac{-8-15.49}{22};x=\frac{-8+15.49}{22} \\ x=-1.0678;x=0.340 \end{gathered}[/tex]it is asking me to find the value of x which i’m confused about
Given angles are alternate exteior angles: pair of angles on the outer side of each of parallel lines but opposite sides of the transversal.
When two lines being crossed are parallel line the alternate exterior angles are equal:
[tex]5x-5=6x-27[/tex]Use the equation above to find the value of x:
[tex]\begin{gathered} 5x-5x-5=6x-5x-27 \\ -5=x-27 \\ -5+27=x-27+27 \\ 22=x \\ \\ x=22 \end{gathered}[/tex]Then, the value of x is 22find the measure of angle 7. assume lines L and M are parallel. figure not drawn to scale.
Answer:
Explanation:
Since the lines L and M are paralle, angles 1 and 6 are equal:
Angle 1 = Angle 6 = 60°
We can also notice that angles 6 and 7 supplementary angles because they are in a straight line. In addition, supplementary angles are two angles whose measures add up to 180°. So,
Angle 6 + Angle 7 =180
We plug in what we know:
Angle 6 + Angle 7 =180
60+ Angle 7 =180
Simplify and rearrange
Angle 7=180-60
Angle 7= 120
Select the correct answer. Which is the inverse of this matrix? 1 2 5 35 9 11 -2. A. -19 9 - 71 15 -7 6 -2 1 -199 -2 1 15 -7 -7 6 B. Oc. [152 7 0 ] OD. - 19 15 - 7 -2 1 O E. The matrix is noninvertible.
We will have that the inverse of the matrix will be the following:
[tex]A=\mleft[\begin{array}{ccc}1 & 2 & 5 \\ 3 & 5 & 9 \\ 1 & 1 & -2 \\ \end{array}\mright][/tex][tex]A^{-1}=\mleft[\begin{array}{ccc}-19 & 9 & -7 \\ 15 & -7 & 6 \\ -2 & 1 & -1 \\ \end{array}\mright][/tex]A bag of marbles can be shared equally among 4, 6 or 18 people with none left over. What is the smallest number of marbles that could be in the bag?
We have a bag of marbles with an unknown amount inside. We only now that if we split the number between 4,6 or 18 people, there are none left. This means, mathematically, that the quantity inside is a multiple of 4, 6 and 18. Since we want to calculate the smallest number of marbles that could be inside, this means that we want to calculate the least common multiple of 4, 6 and 18.
To do this, we proceed as follows.
First, we write all three numbers one next to the other.
4 6 18
Now, we will search among the prime numbers, which one divides at least one of them. We keep doing so until we have divided all numbers so we get the number 1 Recall that the list of the first 10 prime numbers is 1,2,3,5,7,11,13,17,19,23.
So, at the beginning 4, 6 and 18 can be all divided by 2 (we keep this number in mind), then we choose 2. If we divide each number by 2, we get
2 3 9
Now, we repeat the process. Since 3 and 9 are multiple of 3, we choose 3. So now we divide 3 and 9 by 3, so we get
2 1 3
Now, we can divide 2 by 2, so we get
1 1 3
Finally, we divide 3 by 3, so we get
1 1 1
So, the numbers we chose the numbers 2,2,3,3. We multiply them all together and we get
[tex]2\cdot2\cdot3\cdot3\text{ = 4}\cdot9\text{ = 36}[/tex]So the smalles number of marbles that could be in the bag is 36.
The sum of three consecutive odd integers is 79. If x represents the first integer, select the equationbelow which coud be used to solve this word problem.O x(x + 1)(x + 2) = 79Ox+2+1+2+3 = 79O2 ++1+2+2 = 79O2 +2 +2 +3 +4 = 79> Next QuestionON
Answer:
The equation that could be used to solve the word problem is;
[tex]x+x+2+x+4=79[/tex]Given that;
The sum of three consecutive odd integers is 79.
And x represent the first integer.
The common difference between consecutive odd integers is;
[tex]\begin{gathered} d=3-1=2 \\ d=2 \end{gathered}[/tex]So, the first, second and third consecutive integers can be represented as;
[tex]x,x+2,x+4[/tex]The sum of the three is;
[tex]x+x+2+x+4=79[/tex]Therefore, the equation that could be used to solve the word problem is;
[tex]x+x+2+x+4=79[/tex]Scores for a common standardized college aptitude test are normally distributed with a mean of 498 and a standard deviation of 115. Randomly selected men are given a Test Preparation Course before taking this test. Assume, for sake of argument, that the preparation course has no effect.If 1 of the men is randomly selected, find the probability that his score is at least 584.9. P(X > 584.9) = Enter your answer as a number accurate to 4 decimal places.If 7 of the men are randomly selected, find the probability that their mean score is at least 584.9. P(M > 584.9) = Enter your answer as a number accurate to 4 decimal places.
Consider X to be the random variable representing the score of any man.
The mean and standard deviation are given to be 498 and 115 respectively,
[tex]\begin{gathered} \mu=498 \\ \sigma=115 \end{gathered}[/tex]The standard normal variate corresponding to any value of score, is given by the formula,
[tex]z=\frac{x-\mu}{\sigma}[/tex]For X=584.9, the z-score becomes,
[tex]\begin{gathered} z=\frac{584.9-498}{115} \\ z=\frac{86.9}{115} \\ z=0.75 \end{gathered}[/tex]Then the probability that a randomly selected man has scored at least 584.9 is calculated as,
[tex]P(X\ge584.9)=P(z\ge0.75)[/tex]Using the properties of normal distribution,
[tex]\begin{gathered} P(X\ge584.9)=P(z\ge0)-P(0\leq z\leq0.75) \\ P(X\ge584.9)=0.5-\varnothing(0.75) \end{gathered}[/tex]From the Standard Normal Distribution Table,
[tex]\varnothing(0.75)=0.2734[/tex]Substitute the value,
[tex]\begin{gathered} P(X\ge584.9)=0.5-0.2734 \\ P(X\ge584.9)=0.2266 \end{gathered}[/tex]Thus, there is a 0.2266 probability that the score of a randomly selected man is at least 584.9.
Consider a normal sample of 7 men from the college,
[tex]n=7[/tex]The z-score for the random sample is given by,
[tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt[]{n}}}[/tex]Substitute the values,
[tex]\begin{gathered} z=\frac{584.9-498}{\frac{115}{\sqrt[]{7}}} \\ z=\frac{86.9\sqrt[]{7}}{115} \\ z=2.00 \end{gathered}[/tex]So the probability that the mean of the sample is at least 584.9, is calculated as,
[tex]P(X_s\ge584.9)=P(z\ge2)[/tex]Using the properties of Normal Distribution,
[tex]\begin{gathered} P(X\ge584.9)=P(z\ge0)-P(0From the Standard Normal Distribution Table,[tex]\varnothing(2)=0.4772[/tex]Substitute the value,
[tex]\begin{gathered} P(X_s\ge584.9)=0.5-0.4772 \\ P(X_s\ge584.9)=0.0228 \end{gathered}[/tex]Thus, there is a 0.0228 probability that their mean score is at least 584..9.
Which set of ordered pairs represents y as a function of x?O {(3, 3), (3,4), (4,3), (4,4)}O {(2, -1), (4, -2), (6,-3), (8, -4)}|O {(0,0), (1, 1), (1, 0), (2, 1)}o {(1, -5), (1,5), (2, -10), (2, -15)
Answer:
Second choice from the top.
Explanation.
An ordered pair will be considered a function if no two different values of y are given by the same value of x.
For example, a set of ordered pairs constaining (4, 3) and (4, 8) would not be considered a function.
Looking at the choices given we see that only the second choice in the couln gives
You are making the kite shown at the right from five pairs of congruent panels. In parts (a)–(d) below, use the given information to find the side lengths of the kite’s panels. ABCD is a kite. EB = 15 in., BC = 25 in. The extended ratio XY: YZ : ZC is 3: 1: 4. EX is parallel to BC, EX is parallel to YF is parallel to GZ a. nBEX c. YFGZ b. XEFY d. nZGC
Answer:
a. Triange BEX = 9 Inches
b. XEFY = 6 Inches
c. YFGZ = 2Inches
d. Triangle ZGC = 8 Inches
Explanation:
Given that:
BC = 25 Inch
EB=15 Inch
(a)In Triangle BEX
EB is the Hypotenuse.
Using the idea of Pythagorean Triples (9,12,15), the other two legs of Triangle BEX will be 12 Inches and 9 Inches respectively.
From observation, BX is shorter than EX, therefore:
Side Length BX = 9 Inches.
BC=BX+XC
25=9+XC
XC=25-9
XC=16 Inches.
Given that the extended ratio:
[tex]XY\colon YZ\colon ZC=3\colon1\colon4.[/tex][tex]\begin{gathered} XY=\frac{3}{8}\times16=6\text{ Inches} \\ YZ=\frac{1}{8}\times16=2\text{ Inches} \\ ZC=\frac{4}{8}\times16=8\text{ Inches} \end{gathered}[/tex]Therefore, the side lengths of the given panels are:
• a. Triange BEX = 9 Inches
,• b. XEFY = 6 Inches
,• c. YFGZ = 2Inches
,• d. Triangle ZGC = 8 Inches
Part B of this question please help all work shown thank you!
Solution
The vector u = PQ has initial point P(2, 14) and Q(7,3) and
The vector v = RS has initial point R(29, 8) and (12, 17)
For linear form
[tex]y_2-y_1,x_2-x_1[/tex][tex]\begin{gathered} \bar{U}=(7-2,3-14) \\ =(5,-11) \\ \bar{V}=(12-29,17-8) \\ =(-17,9) \end{gathered}[/tex]Part A
[tex]\begin{gathered} \bar{U}=5i-11j \\ \bar{V}=-17i+9j \end{gathered}[/tex]Part B
[tex]\begin{gathered} \theta=\tan ^{-1}(-\frac{11}{5})=-65.56^0 \\ \bar{U}=(\sqrt[]{146}\cos (-65.56),\sqrt[]{146}\sin (-65.56) \\ \gamma=\tan ^{-1}(-\frac{9}{17})+180^0=152.10^0 \\ \bar{V}=(\sqrt[]{370}\cos (152.1),\sqrt[]{370}sin(152.1) \end{gathered}[/tex]Part C
[tex]\begin{gathered} 7\bar{U}-4\bar{V} \\ 7(5,-11)-4(-17,9) \\ (35,-77)-(-68,36) \\ (35+68,-77-36) \\ (103,-113 \\ so \\ 7\bar{U}-4\bar{V}=(103,-113) \end{gathered}[/tex]Jessie has 14 rocks and 2 bags. She wants to place an equal number of rocks in each bag.How many rocks should Jessie put in each bag?
To place an equal number of rocks in each bag we must divide 14 by 2, the result of that division will give how many rocks should be in each bag, then
Therefore
[tex]14\div2=7[/tex]Then Jessie should put 7 rocks in each bag
Find the side labeled x. (Round your answer to one decimal place.) 23 17 1080
x = 15.1 (1 decimal place)
Explanation:we apply the cosine rule:
a² = b² + c² -2bc CosA
let a = 23, b = 17, c = x
a is the side facing the given angle
CosA = 108°
23² = 17² + x² -2(17)(x)Cos 108
529 = 289 + x²- 34(Cos108)
subtract 289 from both sides:
529 - 289 = x²- 34(Cos108)
240 = x²- 34(Cos108)
240 + 34(Cos108) = x²
240 + 34(-0.3090) = x²
240 - 10.506 = x²
229.494 = x²
Square root both sides:
√229.494 = √x²
x = 15.149
x = 15.1 (1 decimal place)
For each ordered pair (x,y) determine whether it is a solution to the inequality y<4
We have an ordered pair (x,y) which we must evaluate for the inequality:
[tex]y<4[/tex]To solve this, we have to replace the variable "y" in the inequality and check if it is true or not.
First-order pair
[tex]\begin{gathered} (4,7) \\ y<4 \\ 7>4 \end{gathered}[/tex]This is not a solution to inequality.
Second-order pair
[tex]\begin{gathered} (5,-4) \\ y<4 \\ -4<4 \end{gathered}[/tex]This is a solution to inequality.
Third-order pair
[tex]\begin{gathered} (-8,4) \\ y<4 \\ 4=4 \end{gathered}[/tex]This is not a solution to inequality.
Fourth-order pair
[tex]\begin{gathered} (-3,9) \\ y<4 \\ 9>4 \end{gathered}[/tex]This is not a solution to inequality.
Finally, we have that the only solution to inequality is:
[tex](5,-4)[/tex]In January, Joanna deposited $250into her savings account. InFebruary, she deposited anadditional $100. If her account hasan APR of 6% compounded monthly,how much interest did Joanna earnin the first two months?
Given:
In January Joanna deposited $250 into her savings account.
In February, she deposited an additional $100.
Her account has an APR of 6% compounded monthly.
Required:
We have to find how much interest did Joanna earn in the first two months.
Explanation:
For the month of January:
[tex]A=P(1+\frac{r}{100})^n[/tex]Here, P=$250, r=6%, amd n= 1 month=1/12 year.
Then,
[tex]\begin{gathered} A=250(1+\frac{6}{100})^{\frac{1}{12}} \\ \\ A=250(\frac{106}{100})^{\frac{1}{12}} \end{gathered}[/tex][tex]\begin{gathered} A=250\times1.005 \\ A=\text{ \$}251.25 \end{gathered}[/tex]Then the interest is
[tex]I=A-P=251.25-250=\text{ \$}1.25[/tex]For the month of February:
P=251.24+100=351.25
Then we have
[tex]\begin{gathered} A=351.25(1+\frac{6}{100})^{\frac{1}{12}} \\ \\ A=351.25(\frac{106}{100})^{\frac{1}{12}} \end{gathered}[/tex][tex]\begin{gathered} A=351.5\times1.005 \\ A=\text{ \$}353.01 \end{gathered}[/tex]Then the interest is
[tex]A=P-I=353.01-351.25=\text{ \$}1.76[/tex]Therefore, the total interest is
[tex]1.25+1.76=\text{ \$}3.01[/tex]Final answer:
Hence the final answer is
[tex]\text{ \$}3.01[/tex]
Select the domain and the range for the graphed function below(4.)
Given:
The objective is to select the correct domain and range of the given graph.
Explanation:
To find domain:
Domain is defined as all the possible values of x in the graph. By observing the graph, the end points are continuing all the way from negative x-axis to positive x-axis.
Then, the domain of the graph will be all real numbers.
[tex]Domain\colon\text{ (-}\infty,+\infty)[/tex]To find range:
The range is defined as the all possible values of y in the graph.
By considering the graph, the vertex lies on the coordinates (-3,3).
For a parabolic graph,
If a < 0, the range will be y ≤ 3.
If a > 0, the range will be y ≥ 3.
Since, the parabolic graph is open downwards then a must be less than zero.
Thus, the range of the graph will be all real numbers less than or equal to 3.
[tex]\text{Range: }(-\infty,3\rbrack[/tex]Hence,
Domain: All real numbers.
Range: All real numbers ≤ 3.
A physical education teacher plans to divide the seventh graders at Wilson middle school into teams of equal size for a year-ending mock Olympic event. He wants each team to have between 4 and 8 students, and all teams need to have the same number of students. The seventh grade of Wilson consists of three classes; one with 28 student, one with 29, and one with 34. How many students should be be on each team?
Answer:
• It is possible to divide the seventh graders into teams of equal sizes.
,• 13 students should be on each team.
Explanation:
The seventh grade of Wilson consists of three classes; one with 28 students, one with 29, and one with 34. Therefore, the total number of students in seventh grade is:
[tex]\text{Total}=28+29+34=91[/tex]He wants each team to have between 4 and 8 students.
[tex]\begin{gathered} \frac{91}{4}\approx22.75 \\ \frac{91}{5}\approx18.2 \\ \frac{91}{6}\approx15.2 \\ \frac{91}{7}\approx13 \\ \frac{91}{8}\approx11.4 \end{gathered}[/tex]Therefore:
• It is possible to divide the seventh graders into teams of equal sizes.
,• 13 students should be on each team.
The salaries of several employees at Company X are listed below:$52000 $62500 $56000$49000$67500 $60000What is the median salary of the employees at Company X?
The median salary of the employees at company X is $58000
Explanation:Given:
The median salaries: $52000, $62500, $56000, $49000, $67500, $60000
To find:
the median salary of the employees
To determine the median, we need to rearrange the salaries. Then, we will pick the median salary
$49000, $52000, $56000, $60000, $62500, $67500
There are 6 salaries. The median salary will be the sum of the two middle salaries divided by 2
[tex]\begin{gathered} median\text{ = }\frac{56000\text{ + 60000}}{2} \\ \\ median\text{ = }\frac{116000}{2} \\ \\ median\text{ = 58000} \end{gathered}[/tex]The median salary of the employees at company X is $58000
going to send you a pic of my question
We know that we can have
0, 1, 2,... pencils
This mean that we cannot have a negative number of pencils (that would not have sense)
Then choice C and D are discarded
Since we can have 8 or less and we could have 0 pencils then the options are
0, 1, 2, 3, 4, 5, 6, 7 or 8
Answer: A