Answers
Answer:
D. 2 (3a + 6)
Step-by-step explanation:
Your friend is correct because when you multiply by 2 inside the bracket you get 6a + 12 which means it is equivalent to the expression in problem.
Related Questions
Proofs are used to show that a mathematical statement is true. The most common form of mathematical statements are if-then statements. Give an example of a true mathematical statement and a false mathematical statement in if-then form. For the false statement, include a counterexample showing that the statement isn’t true.
Answers
Answer:
True mathematical statement.
"If x = 0, then for any real number y, we have: y*x = 0."
This is true, and we can prove it with the axioms of the real set.
A false mathematical statement can be:
"if n and x are integer numbers, then n/x is also an integer number."
And a counterexample of this is if we took n = 1 and x = 2, both are integer numbers, so the first part is true, but:
n/x = 1/2 = 0.5 is not an integer number, then the statement is false,
Answer:
True mathematical statement.
"If x = 0, then for any real number y, we have: y*x = 0."
This is true, and we can prove it with the axioms of the real set.
A false mathematical statement can be:
"if n and x are integer numbers, then n/x is also an integer number."
And a counterexample of this is if we took n = 1 and x = 2, both are integer numbers, so the first part is true, but:
n/x = 1/2 = 0.5 is not an integer number, then the statement is false,
Step-by-step explanation:
compute the missing data in the table for the following exponential function f(x)={1/4}
Answers
Answer:
1/256
Step-by-step explanation:
The table shows a chain of fractions for f(x), x1 is 1/4, x2 is 1/16 and x3 is 1/64. All you need to do is multiply the denominator by 4 and put 1 over it. 64*4 = 256, adding the 1 as the numerator gives us the answer of 1/256 as x4.
If C(x) = 16000 + 600x − 1.8x2 + 0.004x3 is the cost function and p(x) = 4200 − 6x is the demand function, find the production level that will maximize profit. (Hint: If the profit is maximized, then the marginal revenue equals the marginal cost.)
Answers
Answer:
Quantity that will maximize profit=1000
Step-by-step explanation:
Assume quantity=x
Revenue=price*quantity
=(4200-6x)x
=4200x-6x^2
Marginal revenue(MR) =4200-12x
Cost(x)= 16000 + 600x − 1.8x2 + 0.004x3
Marginal cost(MC) =600-3.6x+0.012x^2
Marginal cost=Marginal revenue
600-3.6x+0.012x^2=4200-12x
600-3.6x+0.012x^2-4200+12x=0
0.012x^2-8.4x-3600=0
Solve the quadratic equation using
x= -b +or- √b^2-4ac/2a
a=0.012
b=-8.4
c=-3600
x=-(-8.4) +or- √(-8.4)^2- (4)(0.012)(-3600) / (2)(0.012)
= 8.4 +or- √70.56-(-172.8) / 0.024
= 8.4 +or- √70.56+172.8 / 0.024
= 8.4 +or- √243.36 / 0.024
= 8.4 +or- 15.6/0.024
= 8.4/0.024 +15.6/0.024
= 350+650
x=1000
OR
= 8.4/0.024 -15.6/0.024
= 350 - 650
= -300
x=1000 or -300
Quantity that maximises profits can not be negative
So, quantity that maximises profits=1000
How do I solve this problem
Answers
Answer:
It would take 1 more mile if he took route Street A and then Street B rather than just Street C.
Step-by-step explanation:
Pythagorean Theorem: a² + b² = c²
We use the Pythagorean Theorem to find the length of Street C:
2² + 1.5² = c²
c = √6.25
c = 2.5
Now we find how much longer route A and B is compared to C:
3.5 - (2 + 1.5) = 3.5 - 2.5 = 1
Let the test statistic T have a t distribution when H0 is true. Give the significance level for each of the following situation.
A. Ha: mu > m0, df = 15, rejection region t > 3.733
B. Ha : mu < mu 0, n = 24, rejection region t < - 2.500
C. Ha: mu not = mu 0, n = 31, rejection region t >1.697 or t < - 1.697
Answers
Answer:
a) The degrees of freedom are given by:
[tex]df = 15[/tex]
And the rejection region is [tex]t_{\alpha}<3.733[/tex]
And the significance level would be:
[tex]P(t_{15} >3.733) =0.001[/tex]
b) The degrees of freedom are given by:
[tex]df = 24-1=23[/tex]
And the rejection region is [tex]t_{\alpha} <-2.5[/tex]
And the significance level would be:
[tex]P(t_{23} <-2.5) =0.0099 \approx 0.01[/tex]
c) The degrees of freedom are given by:
[tex]df = 31-1=30[/tex]
And the rejection region is [tex]t_{\alpha} <-1.697[/tex] or [tex]t_{\alpha} >1.697[/tex]
And the significance level would be:
[tex]2*P(t_{30} <-1.697) =0.10[/tex]
Step-by-step explanation:
Part a
We have the following system of hypothesis:
Null hypothesis: [tex]\mu \leq \mu_0 [/tex]
Alternative hypothesis: [tex]\mu > \mu_0 [/tex]
The degrees of freedom are given by:
[tex]df = 15[/tex]
And the rejection region is [tex]t_{\alpha}<3.733[/tex]
And the significance level would be:
[tex]P(t_{15} >3.733) =0.001[/tex]
Part b
We have the following system of hypothesis:
Null hypothesis: [tex]\mu \geq \mu_0 [/tex]
Alternative hypothesis: [tex]\mu < \mu_0 [/tex]
The degrees of freedom are given by:
[tex]df = 24-1=23[/tex]
And the rejection region is [tex]t_{\alpha} <-2.5[/tex]
And the significance level would be:
[tex]P(t_{23} <-2.5) =0.0099 \approx 0.01[/tex]
Part c
We have the following system of hypothesis:
Null hypothesis: [tex]\mu = \mu_0 [/tex]
Alternative hypothesis: [tex]\mu \neq \mu_o[/tex]
The degrees of freedom are given by:
[tex]df = 31-1=30[/tex]
And the rejection region is [tex]t_{\alpha} <-1.697[/tex] or [tex]t_{\alpha} >1.697[/tex]
And the significance level would be:
[tex]2*P(t_{30} <-1.697) =0.10[/tex]
If a ball is thrown into the air with a velocity of 43 ft/s, its height (in feet) after t seconds is given by y = 43t - 16t^2. Find the velocity when t = 2.
Let h = s(t) = 50t − 16t2 give the height of the ball at time t. Then the ball's velocity at time t = a can be found by v(a) = lim t→a s(t) − s(a) t
Answers
Answer:
h(2) = 22 ft/s
Step-by-step explanation:
Simply plug in 2 for t:
h(2) = -16(2)² + 43(2)
h(2) = -16(4) + 86
h(2) = -64 + 86
h(2) = 22
PLZZZZ HELP! BRAINLIEST :)
Answers
Answer: An exponential function
Step-by-step explanation:
The data curves upward, slowly becoming closer and closer to a vertical line.
Hope it helps <3
Find sets of parametric equations and symmetric equations of the line that passes through the two points (if possible). (For each line, write the direction numbers as integers.) (0, 0, 25), (10, 10, 0)
Answers
Answer:
a)Parametric equations are
X= -10t
Y= -10t and
z= 25+25t
b) Symmetric equations are
(x/-10) = (y/-10) = (z- 25)/25
Step-by-step explanation:
We were told to fin two things here which are ; a) the parametric equations and b) the symmetric equations
The given two points are (0, 0, 25)and (10, 10, 0)
The direction vector from the points (0, 0, 25) and (10, 10, 0)
(a,b,c) =( 0 -10 , 0-10 ,25-0)
= < -10 , -10 ,25>
The direction vector is
(a,b,c) = < -10 , -10 ,25>
The parametric equations passing through the point (X₁,Y₁,Z₁)and parallel to the direction vector (a,b,c) are X= x₁+ at ,y=y₁+by ,z=z₁+ct
Substitute (X₁ ,Y₁ ,Z₁)= (0, 0, 25), and (a,b,c) = < -10 , -10 ,25>
and in parametric equations.
Parametric equations are X= 0-10t
Y= 0-10t and z= 25+25t
Therefore, the Parametric equations are
X= -10t
Y= -10t and
z= 25+25t
b) Symmetric equations:
If the direction numbers image and image are all non zero, then eliminate the parameter image to obtain symmetric equations of the line.
(x-x₁)/a = (y-y₁)/b = (z-z₁)/c
CHECK THE ATTACHMENT FOR DETAILED EXPLANATION
The function y=3/-x-3 is graphed only over the domain of{x|-8
Answers
Full Question:
The function [tex]y =\sqrt[3]{-x} - 3[/tex] is graphed only over the domain of {x | –8 < x < 8}. what is the range of the graph?
Answer:
[tex]-5 < y < -1[/tex]
Step-by-step explanation:
Given
Function:[tex]y =\sqrt[3]{-x} - 3[/tex]
Range: {x | –8 < x < 8}
Required
Find the range of the graph
To calculate the range of the graph; we simply substitute the value of x (the domain) at both ends to the given function;
In other words, solve for y when x = -8 and when x = 8
To start with;
When x = -8
[tex]y =\sqrt[3]{-x} - 3[/tex]
[tex]y =\sqrt[3]{-(-8)} - 3}[/tex]
[tex]y =\sqrt[3]{8} - 3}[/tex]
[tex]y =2 - 3[/tex]
[tex]y = -1[/tex]
When x = 8
[tex]y =\sqrt[3]{-8} - 3[/tex]
[tex]y =-2 - 3[/tex]
[tex]y = -5[/tex]
Converting both values of y to inequalities
[tex]-5 < y < -1[/tex]
Hence, the range of the graph is [tex]-5 < y < -1[/tex]
A researcher tests five individuals who have seen paid political ads about a particular issue. These individuals take a multiple-choice test about the issue in which people in general (who know nothing about the issue) usually get 40 questions correct. The number correct for these five individuals was 48, 41, 40, 51, and 50. Using the .05 level of significance, two-tailed, do people who see the ads score differently on this test
Use steps of hypothesis testing sketch distribution involved
Answers
Answer:
[tex]t=\frac{46-40}{\frac{5.148}{\sqrt{5}}}=2.606[/tex]
The degrees of freedom are given by:
[tex]df=n-1=5-1=4[/tex]
The p value wuld be given by:
[tex]p_v =2*P(t_{(4)}>2.606)=0.060[/tex]
For this case the p value is higher than the significance level so then we can conclude that the true mean is not significantly different from 40
The distribution with the critical values are in the figure attached
Step-by-step explanation:
Information given
48, 41, 40, 51, and 50
The sample mean and deviation can be calculated with these formulas:
[tex]\bar X= \frac{\sum_{i=1}^n X_i}{n}[/tex]
[tex]s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}[/tex]
[tex]\bar X=46[/tex] represent the mean height for the sample
[tex]s=5.148[/tex] represent the sample standard deviation
[tex]n=5[/tex] sample size
[tex]\mu_o =40[/tex] represent the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test
Hypothesis to test
We want to test if the true mean for this case is equal to 40, the system of hypothesis would be:
Null hypothesis:[tex]\mu = 40[/tex]
Alternative hypothesis:[tex]\mu \neq 40[/tex]
The statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
Replacing we got:
[tex]t=\frac{46-40}{\frac{5.148}{\sqrt{5}}}=2.606[/tex]
The degrees of freedom are given by:
[tex]df=n-1=5-1=4[/tex]
The p value wuld be given by:
[tex]p_v =2*P(t_{(4)}>2.606)=0.060[/tex]
For this case the p value is higher than the significance level so then we can conclude that the true mean is not significantly different from 40
The distribution with the critical values are in the figure attached
PLEASE HELP!
Fill in the reason for statement 3 in proof below:
SAS
AA
SSS
Answers
Answer:
SAS
Step-by-step explanation:
ΔABD ~ ΔECD is similar through:
S - because ED = CD (Given)
A - same angle ∠D (Statement 2)
S - because AD = BD (Given)
Cheers!
Answer:
SAS
Step-by-step explanation:
You can notice that you have ED/AB = CD/BD You have one common angle15 3/4% is equal to which decimal?
Answers
Answer:
2/4%
Step-by-step explanation:
A research assistant sent out a survey to n people, hoping to get as many responses back as possible.
If the number of people who did not respond to the survey was 300 less than the number of people
who did respond, what fraction of the people who received the survey did respond?
Answers
Answer:
[tex]\frac{n+300}{2n}[/tex] is the correct answer.
Step-by-step explanation:
Given that total number of people = n
Let the number of people who responded to the survey = x
Let the number of people who did not respond to the survey = y
[tex]x+y=n ...... (1)[/tex]
As per question statement:
The number of people who did not respond to the survey was 300 less than the number of people who did respond.
i.e. [tex]x-y =300[/tex] ...... (2).
We need to solve the equations (1) and (2).
Adding (1) and (2):
[tex]2x=n+300\\\Rightarrow x = \dfrac{n+300}{2}[/tex]
The fraction of people who received the survey did respond:
[tex]\dfrac{\text{Number of people who responded}}{\text{Total number of people who received the survey}}\\[/tex]
So, the answer is:
[tex]\dfrac{x}{n}[/tex]
Putting the values of x, we get the answer as:
[tex]\dfrac{n+300}{2n}[/tex]
The two-way table shows the medal count for the top-performing countries in the 2012 Summer Olympics. A 5-column table has 5 rows. The first column has entries United notes, China, Russia, Great Britain, Total. The second column is labeled Gold with entries 46, 38, 24, 29, 137. The third column is labeled Silver with entries 29, 27, 26, 17, 99. The fourth column is labeled Bronze with entries 29, 23, 32, 19, 103. The fifth column is labeled Total with entries 104, 88, 82, 65, 339. Which statement is true?
Answers
Which statement is true?
The probability that a randomly selected silver medal was awarded to Great Britain is StartFraction 17 Over 99 EndFraction. The probability that a randomly selected medal won by Russia was a bronze medal is StartFraction 32 Over 103 EndFraction. The probability that a randomly selected gold medal was awarded to China is StartFraction 88 Over 137 EndFraction. The probability that a randomly selected medal won by the United States was a silver medal is StartFraction 104 Over 339 EndFraction.Answer:
(A)The probability that a randomly selected silver medal was awarded to Great Britain is 17/99.
Step-by-step explanation:
The table is given below:
[tex]\left|\begin{array}{l|c|c|c|c|c} &Gold&Silver & Bronze &Total\\United States &46 & 29 & 29 & 104\\China & 38 & 27 & 23 & 88\\Russia & 24 & 26 & 32 &82\\Great Britain & 29 & 17 & 19 & 65\\&&&&&\\Total &137 & 99 & 103 & 339\end{array}\right[/tex]
We calculate the probabilities given in the statements.
(A) The probability that a randomly selected silver medal was awarded to Great Britain
= 17/99
(B)The probability that a randomly selected medal won by Russia was a bronze medal
=32/82
(C)The probability that a randomly selected gold medal was awarded to China
=38/137
(D)The probability that a randomly selected medal won by the United States was a silver medal
=29/104
We can see that only the first statement is true.
Answer: A. The probability that a randomly selected silver medal was awarded to Great Britain is 17/99.
Step-by-step explanation:
I got it right on edge
what is 3/5 of 1800
Answers
Answer:
1080
Step-by-step explanation:
first do 3 times 1800, because they are both the numerators. Then divide that number, which is 5400, by the denominator: 5. You will get 1080.
How do you pronounce B"?
Answers
Answer:
bee
Step-by-step explanation:
4. Four numbers with mean 7.5, mode 6 and median 7 for grade 7
Answers
Answer:
Number set in order: 6, 6, 8, 10
Step-by-step explanation:
Let the four numbers be represented by a, b, c and d;
Mean = ¹/₄(a + b + c + d) = 7.5
a + b + c + d = 30
Mode is 6, which, by extension, means there is a mode, thus 6 must be the most common number;
This has to mean at least 2 of the numbers are 6;
With the median being 7, we know:
¹/₂(b + c) = 7
b + c = 14
b has to be 6, meaning c is 8;
a must also be 6, making d equal to 10.
Need help with this as soon as possible
Answers
Answer:
[tex] 6 + 5\sqrt{5} [/tex]
irrational
cannot
does not
Step-by-step explanation:
[tex] 5\sqrt{5} + 2\sqrt{9} = [/tex]
[tex] = 5\sqrt{5} + 2 \times 3 [/tex]
[tex] = 5\sqrt{5} + 6 [/tex]
[tex] = 6 + 5\sqrt{5} [/tex]
Answer:
See below.
Step-by-step explanation:
First, find the sum:
[tex]5\sqrt{5}+2\sqrt{9}=5\sqrt{5}+2(3)=5\sqrt{5} +6[/tex]
As long as part of the equation is irrational, the entire answer will also be irrational. Thus:
The result is irrational because it cannot be written [as] the ratio of two integers and its decimal expansion does not terminate or repeat.
A rectangular box has a base that is 4 times as long as it is wide. The sum of the height and the girth of the box is 200 feet. (a) Express the volume V of the box as a function of its width w. Determine the domain of V (w).
Answers
Answer:
(a) [tex]V = (-8W^3 + 800W^2)/3[/tex]
(b) [tex]W > 100[/tex]
Step-by-step explanation:
Let's call the length of the box L, the width W and the height H. Then, we can write the following equations:
"A rectangular box has a base that is 4 times as long as it is wide"
[tex]L = 4W[/tex]
"The sum of the height and the girth of the box is 200 feet"
[tex]H + (2W + 2H) = 200[/tex]
[tex]2W + 3H = 200 \rightarrow H = (200 - 2W)/3[/tex]
The volume of the box is given by:
[tex]V = L * W * H[/tex]
Using the L and H values from the equations above, we have:
[tex]V = 4W * W * (200 - 2W)/3[/tex]
[tex]V = (-8W^3 + 800W^2)/3[/tex]
The domain of V(W) is all positive values of W that gives a positive value for the volume (because a negative value for the volume or for the width doesn't make sense).
So to find where V(W) > 0, let's find first when V(W) = 0:
[tex](-8W^3 + 800W^2)/3 = 0[/tex]
[tex]-8W^3 +800W^2 = 0[/tex]
[tex]W^3 -100W^2 = 0[/tex]
[tex]W^2(W -100) = 0[/tex]
The volume is zero when W = 0 or W = 100.
For positive values of W ≤ 100, the term W^2 is positive, but the term (W - 100) is negative, then we would have a negative volume.
For positive values of W > 100, both terms W^2 and (W - 100) would be positive, giving a positive volume.
So the domain of V(W) is W > 100.
The following chart represents the record low temperatures recorded in Phoenix for April-November. Select the answer below that best describes the mean and the median of the data set (round answers to the nearest tenth). A graph titled Phoenix Low Temperatures has month on the x-axis and temperature (degrees Fahrenheit) on the y-axis. April, 32; May, 40; June, 50; July, 61; August, 60; September, 47; October, 34; November, 25. a. The mean is 43.5°F, and the median is 43.6°F. b. The mean is 60.5°F, and the median is 60.5°F. c. The mean is 60°F, and the median is 61°F. d. The mean is 43.6°F, and the median is 43.5°F.
Answers
Answer:
d. The mean is 43.6°F, and the median is 43.5°F.
Step-by-step explanation:
Hello!
The data corresponds to the low temperatures in Phoenix recorded for April to November.
April: 32ºF
May: 40ºF
June: 50ºF
July: 61ºF
August: 60ºF
September: 47ºF
October: 34ºF
November: 25ºF
Sample size: n= 8 months
The mean or average temperature of the low temperatures in Phoenix can be calculated as:
[tex]\frac{}{X}[/tex]= ∑X/n= (32+40+50+61+60+47+34+25)/8= 43.625ºF (≅ 43.6ºF)
The Median (Me) is the value that separates the data set in two halves, first you have to calculate its position:
PosMe= (n+1)/2= (8+1)/2= 4.5
The value that separates the sample in halves is between the 4th and the 5th observations, so first you have to order the data from least to greatest:
25; 32; 34; 40; 47; 50; 60; 61
The Median is between 40 and 47 ºF, so you have to calculate the average between these two values:
[tex]Me= \frac{(40+47)}{2} = 43.5[/tex] ºF
The correct option is D.
I hope this helps!
Answer:
it is d
Step-by-step explanation:
An industrial psychologist conducted an experiment in which 40 employees that were identified as "chronically tardy" by their managers were divided into two groups of size 20. Group 1 participated in the new "It's Great to be Awake!" program, while Group 2 had their pay docked. The following data represent the number of minutes that employees in Group 1 were late for work after participating in the program.
Does the probability plot suggest that the sample was obtained from a population that is normally distributed? Provide TWO reasons for your classification.
Answers
Answer:
The probability plot of this distribution shows that it is approximately normally distributed..
Check explanation for the reasons.
Step-by-step explanation:
The complete question is attached to this solution provided.
From the cumulative probability plot for this question, we can see that the plot is almost linear with no points outside the band (the fat pencil test).
The cumulative probability plot for a normal distribution isn't normally linear. It's usually fairly S shaped. But, when the probability plot satisfies the fat pencil test, we can conclude that the distribution is approximately linear. This is the first proof that this distribution is approximately normal.
Also, the p-value for the plot was obtained to be 0.541.
For this question, we are trying to check the notmality of the distribution, hence, the null hypothesis would be that the distribution is normal and the alternative hypothesis would be that the distribution isn't normal.
The interpretation of p-valies is that
When the p-value is greater than the significance level, we fail to reject the null hypothesis (normal hypothesis) and but if the p-value is less than the significance level, we reject the null hypothesis (normal hypothesis).
For this distribution,
p-value = 0.541
Significance level = 0.05 (Evident from the plot)
Hence,
p-value > significance level
So, we fail to reject the null or normality hypothesis. Hence, we can conclude that this distribution is approximately normal.
Hope this Helps!!!
A class of 30 music students includes 13 who play the piano, 15 who play the guitar, and 9 who play both the piano and the guitar. How many students in the class play neither instrument?
Answers
Answer: 2
Step-by-step explanation:
As given, out of 30 students, 15 play guitar and 13 play piano, thats 28.
Among these, 9 play both the guitar and the piano.
That means, only 2 remaining students play neither instrument. (30-15-13)
Given the following functions, evaluate each of the following
Answers
Answer:
[tex](f+g)(5) = 40\\(f-g)(5) = 22\\(f*g)(5) = 279[/tex]
[tex](f/g)(5) = 31/9[/tex]
Step-by-step explanation:
[tex]f(5) = (5)^2+2(5)-4\\f(5) = 25+10-4\\f(5) = 31[/tex]
[tex]g(5) = 5+4\\g(5) = 9[/tex]
[tex](f+g)(5) = 31+9\\(f+g)(5) = 40[/tex]
[tex](f-g)(5) = 31-9\\(f-g)(5) = 22[/tex]
[tex](f*g)(5) = 31*9\\(f*g)(5) = 279\\[/tex]
[tex](f/g)(5) = 31/9[/tex]
The length of a rectangle is 5M more than twice the width and the area of the rectangle is 63M to find the dimension of the rectangle
Answers
Answer:
width = 4.5 m
length = 14 m
Step-by-step explanation:
okay so first you right down that L = 5 + 2w
then as you know that Area = length * width so you replace the length with 5 + 2w
so it's A = (5 +2w) * w = 63
then 2 w^2 + 5w - 63 =0
so we solve for w which equals 4.5 after that you solve for length : 5+ 2*4.5 = 14
In a particular year, the mean score on the ACT test was 19.6 and the standard deviation was 5.2. The mean score on the SAT mathematics test was 546 and the standard deviation was 126. The distributions of both scores were approximately bell-shaped. Round the answers to at least two decimal placesFind the z-score for an ACT score of 26. The Z-score for an ACT score of 26 is ______ .
Answers
Answer:
0.11
Step-by-step explanation:
Let the random variable score, X = 26; mean, ∪ = 19.6; standard deviation, α = 5.2
By comparing P(0≤ Z ≤ 26)
P(Z ≤ X - ∪/α) = P(Z ≤ 26 - 19.6/5.2)
= P(Z ≤ 1.231)
Using Table: P(0 ≤ Z ≤ 1) = 0.39
P(Z > 1) = (0.5 - 0.39) = 0.11
∴ P(Z > 26) = 0.11
A manufacturer of chocolate chips would like to know whether its bag filling machine works correctly at the 418 gram setting. It is believed that the machine is underfilling the bags. A 9 bag sample had a mean of 413 grams with a standard deviation of 20. A level of significance of 0.1 will be used. Assume the population distribution is approximately normal. Is there sufficient evidence to support the claim that the bags are underfilled?
Answers
Answer:
No. At a significance level of 0.1, there is not enough evidence to support the claim that the bags are underfilled (population mean significantly less than 418 g.)
Step-by-step explanation:
This is a hypothesis test for the population mean.
The claim is that the bags are underfilled (population mean significantly less than 418 g.)
Then, the null and alternative hypothesis are:
[tex]H_0: \mu=418\\\\H_a:\mu< 418[/tex]
The significance level is 0.1.
The sample has a size n=9.
The sample mean is M=413.
As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=20.
The estimated standard error of the mean is computed using the formula:
[tex]s_M=\dfrac{s}{\sqrt{n}}=\dfrac{20}{\sqrt{9}}=6.6667[/tex]
Then, we can calculate the t-statistic as:
[tex]t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{413-418}{6.6667}=\dfrac{-5}{6.6667}=-0.75[/tex]
The degrees of freedom for this sample size are:
[tex]df=n-1=9-1=8[/tex]
This test is a left-tailed test, with 8 degrees of freedom and t=-0.75, so the P-value for this test is calculated as (using a t-table):
[tex]\text{P-value}=P(t<-0.75)=0.237[/tex]
As the P-value (0.237) is bigger than the significance level (0.1), the effect is not significant.
The null hypothesis failed to be rejected.
At a significance level of 0.1, there is not enough evidence to support the claim that the bags are underfilled (population mean significantly less than 418 g.)
At her favorite sneakers store Nyeema saved $48 because of a
sale.
If the sneakers normally cost $120. How much did she save?
Answers
Answer:
40%
Step-by-step explanation:
We can find what percent 48 is of 120 by dividing:
48/120 = 0.4 or 40%
So, she saved 40% from the original price.
Fake Question: Should Sekkrit be a moderator? (answer if you can) Real Question: Solve for x. [tex]x^2+3x=-2[/tex]
Answers
Answer:
x = -2 , -1
Step-by-step explanation:
Set the equation equal to 0. Add 2 to both sides:
x² + 3x = -2
x² + 3x (+2) = - 2 (+2)
x² + 3x + 2 = 0
Simplify. Find factors of x² and 2 that will give 3x when combined:
x² + 3x + 2 = 0
x 2
x 1
(x + 2)(x + 1) = 0
Set each parenthesis equal to 0. Isolate the variable, x. Note that what you do to one side of the equation, you do to the other.
(x + 2) = 0
x + 2 (-2) = 0 (-2)
x = 0 - 2
x = -2
(x + 1) = 0
x + 1 (-1) = 0 (-1)
x = 0 - 1
x = -1
x = -2 , -1
~
Answer:
x = -2 OR x = -1
Step-by-step explanation:
=> [tex]x^2+3x = -2[/tex]
Adding 2 to both sides
=> [tex]x^2+3x+2 = 0[/tex]
Using mid-term break formula
=> [tex]x^2+x+2x+2 = 0[/tex]
=> x(x+1)+2(x+1) = 0
=> (x+2)(x+1) = 0
Either:
x+2 = 0 OR x+1 = 0
x = -2 OR x = -1
P.S. Ummmm maybe...... Because he usually reports absurd answers! So, Won't it be better that he could directly delete it. And one more thing! He's Online 24/7!!!!!
Find the remainder when f(x)=2x3−x2+x+1 is divided by 2x+1.
Answers
Step-by-step explanation:
it can be simply done by using remainder theorem.
The National Safety Council (NSC) estimates that off-the-job accidents cost U.S. businesses almost $200 billion annually in lost productivity (National Safety Council, March 2006). Based on NSC estimates, companies with 50 employees are expected to average three employee off-the-job accidents per year. Answer the following questions for companies with 50 employees.a. What is the probability of no off-the-job accidents during a one-year period (to 4 decimals)?b. What is the probability of at least two off-the-job accidents during a one-year period (to 4 decimals)?c. What is the expected number of off-the-job accidents during six months (to 1 decimal)?d. What is the probability of no off-the-job accidents during the next six months (to 4 decimals)?
Answers
Answer:
a. 0.0498 = 4.98% probability of no off-the-job accidents during a one-year period
b. 0.8008 = 80.08% probability of at least two off-the-job accidents during a one-year period.
c. The expected number of off-the-job accidents during six months is 1.5.
d. 0.2231 = 22.31% probability of no off-the-job accidents during the next six months.
Step-by-step explanation:
We have the mean during a period, so we use the Poisson distribution.
Poisson distribution:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
In which
x is the number of sucesses
e = 2.71828 is the Euler number
[tex]\mu[/tex] is the mean in the given interval.
Companies with 50 employees are expected to average three employee off-the-job accidents per year.
This means that [tex]\mu = 3n[/tex], in which n is the number of years.
a. What is the probability of no off-the-job accidents during a one-year period (to 4 decimals)?
This is [tex]P(X = 0)[/tex] when [tex]\mu = 3*1 = 3[/tex]. So
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-3}*3^{0}}{(0)!} = 0.0498[/tex]
0.0498 = 4.98% probability of no off-the-job accidents during a one-year period.
b. What is the probability of at least two off-the-job accidents during a one-year period (to 4 decimals)?
Either there are less than two accidents, or there are at least two. The sum of the probabilities of these events is 1. So
[tex]P(X < 2) + P(X \geq 2) = 1[/tex]
We want [tex]P(X \geq 2)[/tex]. Then
[tex]P(X \geq 2) = 1 - P(X < 2)[/tex]
In which
[tex]P(X < 2) = P(X = 0) + P(X = 1)[/tex]
[tex]P(X = 0) = \frac{e^{-3}*3^{0}}{(0)!} = 0.0498[/tex]
[tex]P(X = 1) = \frac{e^{-3}*3^{1}}{(1)!} = 0.1494[/tex]
[tex]P(X < 2) = P(X = 0) + P(X = 1) = 0.0498 + 0.1494 = 0.1992[/tex]
[tex]P(X \geq 2) = 1 - P(X < 2) = 1 - 0.1992 = 0.8008[/tex]
0.8008 = 80.08% probability of at least two off-the-job accidents during a one-year period.
c. What is the expected number of off-the-job accidents during six months (to 1 decimal)?
6 months is half a year, so [tex]n = 0.5[/tex]
[tex]\mu = 3n = 3*0.5 = 1.5[/tex]
The expected number of off-the-job accidents during six months is 1.5.
d. What is the probability of no off-the-job accidents during the next six months (to 4 decimals)?
This is P(X = 0) when [tex]\mu = 1.5[/tex]. So
[tex]P(X = 0) = \frac{e^{-1.5}*1.5^{0}}{(0)!} = 0.2231[/tex]
0.2231 = 22.31% probability of no off-the-job accidents during the next six months.