Answer:
start at (0,0) and go up 1, right 1, and mark a point and keep doing that
Step-by-step explanation:
If a pool measures 30 feet by 66 feet, what is the scale of the drawing shown? 1 in =
To determine the scale of a drawing, you need the dimensions on the drawing or map as well as the actual dimensions of the object. The scale is then calculated as the ratio of these two sets of dimensions. However, without the dimensions of the drawing in this case, it is not possible to determine the exact scale.
Explanation:The scale of a drawing is a ratio that compares the dimensions of the physical object to the dimensions on the drawing or map. In this case, the physical object is a pool that measures 30 feet by 66 feet. To find the scale, you would need additional information such as the dimensions of the pool in the drawing.
For example, if the pool is drawn as 2 inches by 4.4 inches on the paper, then the scale of the drawing would be 1 inch = 15 feet (for length) and 1 inch = 15 feet (for width), as 30 feet / 2 inches = 15 feet per inch and 66 feet / 4.4 inches = 15 feet per inch respectively.
Without this additional information, it is not possible to provide the exact scale of the drawing.
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The scale for the drawing shown is 1 in = 6 feets
How to determine the Scale of the drawingUsing the parameters given ;
Actual dimension :
length = 66 feets width = 30 feetsDrawing dimension :
Length = 11 feets width = 5 feetsTo calculate the scale of the drawing : Take the ratio of equivalent sides of the two drawings ;
Actual length / Drawing length = Actual width / drawing widthNow we have ;
Scale = 66 / 11 = 6Hence the scale drawing is :
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mackenzie made 33 out of 50 throws. what is the experimental probability that mackenzie will not make the next free throw she attempts
answer:
1.94117647058
Step-by-step explanation:
33/50 her free throw % is 66% at the moment based on how many she missed you'd do 33 / by 17 you get 1.94117647058
What is the range of the data below?
50
60
70
80
90
100
A(22)
B(28)
C(41)
D(42)
Answer:
it d (42)
Step-by-step explanation:
Answer:
d: 42
Step-by-step explanation:
Whats the answer of this question:
Question: 1+1
Answer:
2
Step-by-step explanation:
Answer: 9?
Step-by-step explanation: I think.
Game consoles: A poll surveyed 341 video gamers, and 89 of them said that they prefer playing games on a console, rather than a computer or hand-held device. An executive at a game console manufacturing company claims that less than 28% of gamers prefer consoles. Does the poll provide convincing evidence that the claim is true? Use the =α0.05 level of significance and the P-value method with the TI-84 Plus calculator.
Answer:
Step-by-step explanation:
Hello!
The objective is to test if the population proportion of gamers that prefer consoles is less than 28% as the manufacturer claims.
Of 341 surveyed players, 89 said that they prefer using a console.
The sample resulting sample proportion is p'= 89/341= 0.26
If the company claims is true then p<0.28, this will be the alternative hypothesis of the test.
H₀: p ≥ 0.28
H₁: p < 0.28
α: 0.05
To study the population proportion you have to use the approximation of the standard normal [tex]Z= \frac{p'-p}{\sqrt{\frac{p(1-p)}{n} } }[/tex]≈N(0;1)
[tex]Z_{H_0}= \frac{0.26-0.28}{\sqrt{\frac{0.28*0.72}{341} } }= -0.82[/tex]
This test is one-tailed left, i.e. that you'll reject the null hypothesis to small values of Z, and so is the p-value, you can obtain it looking under the standard normal distribution for the probability of obtaining at most -0.82:
P(Z≤-0.82)= 0.206
Using the p-value approach:
If p-value ≤ α, reject the null hypothesis
If p-value > α, don't reject the null hypothesis
The decision is to not reject the null hypothesis.
Then at a level of 5%, you can conclude that the population proportion of gamers that prefer playing on consoles is at least 28%.
I hope this helps!
Can someone please help me fill this out?
Given:
The height of the the tree from where the tree was broken = 13 ft
The distance from the foot of the tree to the broken top = 36 ft
To find the height of the tree.
Formula
By Pythagoras theorem we get,
h² = l²+b²,
where, h be the hypotenuse
b be the base and
l be the other side of the triangle along the right angle.
Now,
Putting, l =13 and b = 36 we get,
[tex]h^{2} =36^{2} +13^{2}[/tex]
or, [tex]h^{2} = 1296+169[/tex]
or, [tex]h^{2} =1465[/tex]
or, [tex]h=\sqrt{1465}[/tex]
or, [tex]h = 38.3[/tex]
Therefore,
The height of the tree is about = 13+38.3 ft = 51.3 ft
Hence,
The height of the tree is about 51.3 ft.
Find the output, k, when the input, t, is 3.
k = 13t – 2
k=[
S
Answer:
k = 37
Step-by-step explanation:
[tex]k = 13t - 2 \\ \therefore \: k = 13 \times 3 - 2 \\ \therefore \: k = 39 - 2 \\ \huge \red{ \boxed{ \therefore \: k = 37}}[/tex]
Which system of equations has exactly one solution?
A.) 2 x minus 4 y = 8. x + y = 7.
B.) 3 x + y = negative 1. 6 x + 2 y = negative 2.
C.) 3 x + 3 y = 3. Negative 6 x minus 6 y = 3.
D.) 3 x minus 2 y = 4. Negative 3 x + 2 y = 4.
Answer:
A.) 2 x minus 4 y = 8. x + y = 7.
Step-by-step explanation:
A.) 2 x minus 4 y = 8. x + y = 7.
2x-4y = 8
x+y=7
This has one solution
B.) 3 x + y = negative 1. 6 x + 2 y = negative 2.
3x+y = -1
Multiply by 2
6x +2y = -2
This is the second equation so it has infinite solutions
C.) 3 x + 3 y = 3. Negative 6 x minus 6 y = 3.
3x+3y =3
Multiply by -2
-6x-6y = -6
The equation is the same but the constant, which means they are parallel lines, they have zero solutions
D.) 3 x minus 2 y = 4. Negative 3 x + 2 y = 4.
3x -2y = 4
Multiply by -1
-3x +2y = -4
The equation is the same but the constant, which means they are parallel lines, they have zero solutions
Answer:
A.) 2 x minus 4 y = 8. x + y = 7.
Step-by-step explanation:
A.) 2 x minus 4 y = 8. x + y = 7.
2x - 4y = 8
x - 2y = 4
x = 4 + 2y
x + y = 7
4 + 2y + y = 7
3y = 3
y = 1
x = 4 + 2
x = 6
Only solution: (6,1)
Calibrating a scale:
Making sure that the scales used by businesses in the United States are accurate is the responsibility of the National Institute for Standards and Technology (NIST) in Washington, D.C. Suppose that NIST technicians are testing a scale by using a weight known to weigh exactly 1000 grams. The standard deviation for scale reading is known to be σ = 2.3. They weigh this weight on the scale 49 times and read the result each time. The 49 scale readings have a sample mean of x = 999.0 grams. The calibration point is set too low if the mean scale reading is less than 1000 grams.
1. The technicians want to perform a hypothesis test to determine whether the calibration point is set too low. Use the α = 0.01 level of significance and the P-value method with the TI-84 calculator.
Answer:
We conclude that the calibration point is set too low.
Step-by-step explanation:
We are given that NIST technicians are testing a scale by using a weight known to weigh exactly 1000 grams. The standard deviation for scale reading is known to be σ = 2.3. They weigh this weight on the scale 49 times and read the result each time. The 49 scale readings have a sample mean of x = 999.0 grams.
The calibration point is set too low if the mean scale reading is less than 1000 grams.
Let [tex]\mu[/tex] = mean scale reading
So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu \geq[/tex] 1000 grams {means that the calibration point is not set too low}
Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] < 1000 grams {means that the calibration point is set too low}
The test statistics that would be used here One-sample z test statistics as we know about the population standard deviation;
T.S. = [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
where, [tex]\bar X[/tex] = sample mean = 999 grams
[tex]\sigma[/tex] = population standard deviation = 2.3 grams
n = sample of scale readings = 49
So, test statistics = [tex]\frac{999-1000}{\frac{2.3}{\sqrt{49} } }[/tex]
= -3.04
The value of z test statistics is -3.04.
Now, P-value of the test statistics is given by the following formula;
P-value = P(Z < -3.04) = 1 - P(Z [tex]\leq[/tex] 3.04)
= 1 - 0.99882 = 0.00118
Since, the P-value is less than the level of significance as 0.01 > 0.00118, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.
Therefore, we conclude that the calibration point is set too low.
4 vd
6 yd
20 yd
Surface Area =
Determine the intercepts of the line.
y = -7x +3
y-intercept:
t-intercept: (
-intercept:
,
)
2 points
18) Find the volume of the sphere. Round your answer to the nearest
tenth. (Use Pi: 3.14) *
548.45
538.51
534.45
Answer:
Step-by-step explanation:
Find the volume of a sphere 12 yards a crossed
What is the answer for 8b - 2b -4
Simplfied Answer: [tex]6b-4[/tex]
Combine Like Terms
[tex]8b+-2b+-4\\(8b+-2b)+(-4)\\6b+-4[/tex]
Answer:
[tex]8b - 2b - 4 \\ 6b - 4 \\ = 2(3b - 2)[/tex]
In order to estimate the proportion of students that text while driving, a school administrator selects a simple random sample of students from a list of all students at the school who have parking permits. The students are called to the office, one at a time. The administrator asks each student, “Do you text while driving, even though you are not supposed to?” Based on the survey, the administrator estimates that only 2% of students with parking passes text while driving. What potential bias is present in the design of this survey and what is the likely direction of the bias?
Answer:the answer is (A)
Step-by-step explanation:
Because the survey face-to-face, one at a time, with an administrator of the school the students would likely reply that they do not text and drive even if they actually do. This response bias would cause the sample proportion to underestimate the value of the true proportion of all students at this school that text and drive.
The design of the survey contains response bias, which is likely to underestimate the true proportion of students who text while driving due to students' reluctance to admit to prohibited and risky behaviors to an authority figure.
Explanation:The potential bias present in the design of this survey is response bias.
This type of bias occurs when the way a question is asked leads the respondents to give a certain answer
. In this case, by directly asking the students if they engage in the prohibited behavior of texting while driving, there is a likelihood that students may not answer truthfully due to fear of repercussions, the desire to give socially acceptable answers, or reluctance to admit to risky behaviors in a face-to-face setting with a school authority figure.
The likely direction of the bias is that it would underestimate the true proportion of students who text while driving since students might not want to admit to a behavior that is both dangerous and frowned upon while talking to a school administrator.
A stack of four cards contains two red cards and two black cards. I select two cards, one at a time, and do not replace the first card selected before selecting the second card. Consider the events:A = the first card selected is red.B = the second card selected is red.The events A and B are:conditionals.independent.disjoint.None of the answer options are correct.
Final answer:
The events A and B are mutually exclusive. The events A and C are not mutually exclusive. Two events A and B are independent events.
Explanation:
The events A and B are mutually exclusive because they cannot happen at the same time. P(A AND B) = 0.
No, A and C are not mutually exclusive because they can occur at the same time. In fact, C includes all of the outcomes of A; if the card chosen is blue it is also (red or blue). P(A AND C) = P(A) = 20.
Two events A and B are independent events if the knowledge that one occurred does not affect the chance the other occurs. For example, the outcomes of two roles of a fair die are independent events. The outcome of the first roll does not change the probability for the outcome of the second roll. To show two events are independent, you must show only one of the above conditions. If two events are not independent, then we say that they are dependent events. Sampling may be done with replacement or without replacement. With replacement: If each member of a population is replaced after it is picked, then that member has the possibility of being chosen more than once. When sampling is done with replacement, then events are considered to be independent, meaning the result of the first pick will not change the probabilities for the second pick.
A random sample of n = 83 measurements is drawn from a binomial population with probability of success 0.4. Complete parts a through d below.
a. Give the mean and standard deviation of the sampling distribution of the sample proportion, p. The mean of the sampling distribution of p is The standard deviation of the sampling distribution of p is (Round to four decimal places as needed.)
b. Describe the shape of the sampling distribution of p. 0 A The shape of the sampling distribution of p is approximately normal because the sample size is small. The shape of the sampling distribution of p is approximately normal because the sample size is large The shape of the sampling distribution of p is approximately uniform because the sample size is smal ○ C. O D. The shape of the sampling distribution of p is approximately uniform because the sample size is large.
c. Calculate the standard normal z-score corresponding to a value of p=0.41. The standard normal z-score corresponding to a value of p: 041 is . Round to two decimal places as needed.) Finn)
Answer:
a) The mean of the sampling distribution of p is 0.4.
The standard deviation of the sampling distribution of p is 0.0538.
b) The shape of the sampling distribution of p is approximately normal because the sample size is large.
c) z=0.19
Step-by-step explanation:
We have a random sample of size n=83, drawn from a binomial population with proabiliity p=0.4. We have to compute the characteristic of the sampling distribution of the sample proportion.
a) The mean of the sampling distribution is equal to the mean of the distribution p:
[tex]\mu_p=p=0.4[/tex]
The standard deviation of the sampling distribution is:
[tex]\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.4*0.6}{83}}=\sqrt{0.0029}=0.0538[/tex]
b) The shape of sampling distribution with enough sample size tend to be approximately normal. In this case, n=83 is big enough for a binomial distribution.
c) The z-score fof p-0.41 can be calculated as:
[tex]z=\dfrac{p-\mu_p}{\sigma_p}=\dfrac{0.41-0.4}{0.0538}=\dfrac{0.01}{0.0538}=0.19[/tex]
The mean of the sampling distribution is 0.4 and the standard deviation is 0.0538.
How to solve the sampling distribution?From the information given, the sample is drawn from a binomial population with probability of success 0.4. Therefore, the mean is 0.4.
The standard deviation will be:
= [(✓0.4 × ✓0.6) / ✓83)]
= 0.0538
Furthermore, the shape of the sampling distribution of p is approximately normal because the sample size is large.
The standard normal z-score will be:
= (0.41 - 0.4) / 0.0538
= 0.10/0.0538
= 0.19.
Therefore, the standard normal z-score is 0.19.
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A random variable follows the continuous uniform distribution between 160 and 340. Calculate the following quantities for the distribution. a) P(220less than or equalsxless than or equals290) b) P(160less than or equalsxless than or equals250) c) P(xgreater than190) d) What are the mean and standard deviation of this distribution?
Answer:
a) 0.3889
b) 0.5
c) 0.8333
d) The mean is 250 and the standard deviation is 51.96.
Step-by-step explanation:
An uniform probability is a case of probability in which each outcome is equally as likely.
For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.
The probability of finding a value of X higher than x is:
[tex]P(X > x) = 1 - \frac{x - a}{b-a}[/tex]
The probability of finding a value of X between c and d is:
[tex]P(c \leq X \leq d) = \frac{d - c}{b - a}[/tex]
The mean and the standard deviation are, respectively:
[tex]M = \frac{a+b}{2}[/tex]
[tex]S = \sqrt{\frac{b-a}^{2}{12}}[/tex]
A random variable follows the continuous uniform distribution between 160 and 340.
This means that [tex]a = 160, b = 340[/tex]
a)
[tex]P(220 \leq X \leq 290) = \frac{290 - 220}{340 - 160} = 0.3889[/tex]
b)
[tex]P(160 \leq X \leq 250) = \frac{250 - 160}{340 - 160} = 0.5[/tex]
c)
[tex]P(X > 190) = 1 - \frac{190 - 160}{340 - 160} = 0.8333[/tex]
d)
[tex]M = \frac{160 + 340}{2} = 250[/tex]
[tex]S = \sqrt{\frac{340 - 160}^{2}{12}} = 51.96[/tex]
The mean is 250 and the standard deviation is 51.96.
A quadratic function has an equation of f(x) = a(x + 3)2 – 4 and goes through
the point (-2, 2). What is the a-value of the equation?
Answer:
6
Step-by-step explanation:
Put the given point into the equation and solve for a.
f(-2) = a(-2 +3)² -4 = a -4 = 2
a = 6 . . . . . add 4
The a-value in the equation is 6.
Find product (7m+8)(4m +1)
Step-by-step explanation:
[tex](7m + 8)(4m + 1) \\ = 7m(4m + 1) + 8(4m + 1) \\ = 28 {m}^{2} + 7m + 32m + 8 \\ \purple { \bold{= 28 {m}^{2} + 39m + 8}}[/tex]
Answer:
You can use foil or the box method.
Step-by-step explanation:
Sorry it looks so bad :P
find and list the slope and y-intercept of the equation below. -2y=6(2x-2)
[tex]\text{Solve the equation to make it into slope intercept form}\\\\\text{Slope intercept form: y = mx + b}\\\\\text{Solve:}\\\\-2y=6(2x-2)\\\\\text{Distribute the 6 to the variables inside the parenthesis}\\\\-2y=12x-12\\\\\text{Divide both sides by -2}\\\\y=-6x+6\\\\\text{The slope would be -6 and the y intercept would be 6}\\\\\boxed{\text{Slope: -6 y-intercept: 6}}[/tex]
Find the length of the bold arc. Round to the nearest tenth and use 3.14 or the pi button on your calculator for pi.
Answer:
8π yards
Step-by-step explanation:
A circle subtends a total angle of 360 ° from its center.The length of an arc is directly proportional to the angle it subtends from the circle's center. The arc's length can therefore be calculated as:
C=πd=2πr
Where C is circumference, d is diameter and r is radius. Given r as 16 yards then the arc length which is equivalent to circumference is given as
[tex]C=\frac {90}{360}*2*\pi*16=8\pi[/tex]
Answer:
2,144.7 i just answered the other one is wrong
Step-by-step explanation:
Scores on the SAT college entrance test in a recent year were roughly Normal with mean 1012.5 and standard deviation 184.9. You choose an SRS of 90 students and average their SAT scores. If you do this many times, the mean of the average scores you get will be close to
Answer:
By the Central Limit Theorem, the mean of the average scores you get will be close to 1012.5.
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
Mean score of the population:
1012.5
If you do this many times, the mean of the average scores you get will be close to
By the Central Limit Theorem, the mean of the average scores you get will be close to 1012.5.
0.73 as a fraction. please
Answer:
73/100
Step-by-step explanation:
.73 is 73 hundredth which is also 73/100.
Answer:
73/100
Step-by-step explanation:
.73
There are 2 numbers after the decimal, which means the denominator is 1 with 2 zeros after it
73/100
This is as simple as it get since 73 is a prime number
(1 point) Supppose A is an invertible n×n matrix and v⃗ is an eigenvector of A with associated eigenvalue 8. Convince yourself that v⃗ is an eigenvector of the following matrices, and find the associated eigenvalues. The matrix A4 has an eigenvalue . The matrix A−1 has an eigenvalue . The matrix A+4In has an eigenvalue . The matrix 5A has an eigenvalue .
By definition, if [tex]\vec v[/tex] is an eigenvector of [tex]\mathbf A[/tex] (with associated eigenvalue 8), then
[tex]\mathbf A\vec v=8\vec v[/tex]
Notice that[tex]\mathbf A^4\vec v=\mathbf A^3\cdot\mathbf A\vec v=8\mathbf A^3\vec v[/tex]
Rinse and repeat to find
[tex]\mathbf A^4\vec v=8^4\vec v[/tex]
so that [tex]\mathbf A^4[/tex] has a corresponding eigenvalue of 4096.
Let [tex]\lambda[/tex] be the eigenvalue of [tex]\mathbf A^{-1}[/tex] corresponding to [tex]\vec v[/tex]. Then[tex]\mathbf A^{-1}\vec v=\lambda\vec v\implies\mathbf A\cdot\mathbf A^{-1}\vec v=\mathbf A\cdot\lambda \vec v\implies\vec v=8\lambda\vec v[/tex]
For this to hold, we require [tex]8\lambda=1[/tex], or [tex]\lambda=\frac18[/tex]. So [tex]\mathbf A^{-1}[/tex] has a corresonding eigenvalue of 1/8.
Expanding gives[tex](\mathbf A+4\mathbf I_n)\vec v=\mathbf A\vec v+4\mathbf I_n\vec v=8\vec v+4\vec v=12\vec v[/tex]
so that [tex]\mathbf A+4\mathbf I_n[/tex] has an associated eigenvalue of 12.
You know the drill:[tex]5\mathbf A\vec v=5\cdot8\vec v=40\vec v[/tex]
so the eigenvalue of [tex]5\mathbf A[/tex] is 40.
The matrix [tex]A^{4}[/tex] has eigen value of 4096.
The matrix [tex]A^{-1}[/tex] has eigen value of [tex]\frac{1}{8}[/tex]
The eigen value of (A+4I) is 12.
The eigen value of [tex]5Av[/tex] is 40.
Eigen values of matrix:Eigenvalues are the set of scalar values that is associated with the set of linear equations most probably in the matrix equations.
The eigenvectors are also termed as characteristic roots.
If v is an eigen vector of matrix A associated with eigen value [tex]\lambda[/tex].
Then, [tex]Av=\lambda v[/tex]
it is given that, v is an eigen vector of A with eign value 8.
so taht, [tex]Av=8v[/tex]
We have to find eigen value for [tex]A^{4}[/tex].
[tex]A^{4}v=A^{3}(8v)=8^{4}v=4096v[/tex]
Thus, [tex]A^{4}[/tex] has eigen value of 4096.
If A has eigen value of 8 . Then [tex]A^{-1}[/tex] has eigen value of [tex]\frac{1}{8}[/tex]
The eigen value of (A+4I) is computed as,
[tex](A+4I_{n})v=Av+4v=8v+4v=12v[/tex]
The eigen value of (A+4I) is 12.
The eigen value of [tex]5Av[/tex] is,
[tex]5Av=5*8v=40v[/tex]
The eigen value of [tex]5Av[/tex] is 40.
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Type the noun form of the verb.
contribute
Answer:
Contribution
Step-by-step explanation:
The verb would be used as:
I like to contribute to discussions.
The noun would be used as:
My contribution to the discussion was alright.
I hope this helped! (Sorry for the dry examples, I couldn't think of anything else)
A husband and wife, Ed and Rina, share a digital music player that has a feature that randomly selects which song to play. A total of 3476 songs have been loaded into the player, some by Ed and the rest by Rina. They are interested in determining whether they have each loaded different proportions of songs into the player. Suppose that when the player was in the random-selection mode, 35 of the first 53 songs selected were songs loaded by Rina. Let p denote the proportion of songs that were loaded by Rina.
State the null and alternative hypotheses to be tested. How strong is the evidence that Ed and Rina have each loaded a different proportion of songs into the player? Make sure to check the conditions for the use of this test. (Round your test statistic to two decimal places and your P-value to four decimal places. Assume a 95% confidence level.) Hypotheses:A) H0: p = 0.5 Ha: p < 0.5B) H0: p = 0.5 Ha: p ≠ 0.5C) H0: p = 0.5 Ha: p > 0.5
The correct hypotheses to be tested are:
H0: p = 0.5 (The proportion of songs loaded by Rina is 0.5, meaning Ed and Rina loaded an equal proportion of songs.)
Ha: p = 0.5 (The proportion of songs loaded by Rina is not equal to 0.5, meaning Ed and Rina loaded different proportions of songs.)
To determine how strong the evidence is that Ed and Rina have each loaded a different proportion of songs into the player, we can perform a hypothesis test. The test statistic for a proportion in a binomial setting is given by:
[tex]\[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \][/tex]
where:
- [tex]\(\hat{p}\)[/tex] is the sample proportion of successes (in this case, the proportion of songs by Rina),
- [tex]\(p_0\)[/tex] is the hypothesized proportion under the null hypothesis (0.5),
- [tex]\(n\)[/tex] is the sample size (53 songs).
Given that 35 out of 53 songs were by Rina, we calculate [tex]\(\hat{p}\)[/tex] as:
[tex]\[ \hat{p} = \frac{35}{53} \approx 0.660 \][/tex]
Now we can calculate the test statistic [tex]\(z\)[/tex] :
[tex]\[ z = \frac{0.660 - 0.5}{\sqrt{\frac{0.5(1-0.5)}{53}}} = \frac{0.160}{\sqrt{\frac{0.25}{53}}} \approx \frac{0.160}{\sqrt{0.004717}} \approx \frac{0.160}{0.0686} \approx 2.33 \][/tex]
The P-value for a two-tailed test is the probability of observing a test statistic as extreme as or more extreme than the one observed, under the assumption that the null hypothesis is true. We can find the P-value by looking up the z-score in a standard normal distribution table or using a calculator:
[tex]\[ P\text{-value} = 2 \times P(Z > |2.33|) \approx 2 \times 0.0099 \approx 0.0198 \][/tex]
Since the P-value (0.0198) is less than the significance level of 0.05, we reject the null hypothesis in favour of the alternative hypothesis. This means there is strong evidence to suggest that Ed and Rina have each loaded a different proportion of songs into the player.
Before we conclude, we must check the conditions for the use of this test:
1. The samples are independent. Each song selection is made independently of the others.
2. The number of successes and failures are each at least 10. In this case, Rina's songs (35) and Ed's songs (18) are both greater than 10, so this condition is satisfied.
3. The sample size is less than 10% of the population size. The population size is 3476 songs, and the sample size is 53 songs, which is much less than 10% of the population, so this condition is also satisfied.
Since all conditions are met, we can confidently conclude that there is strong evidence that Ed and Rina have each loaded a different proportion of songs into the player.
To solve the system of linear equations 3 x minus 2 y = 4 and 9 x minus 6 y = 12 by using the linear combination method, Henry decided that he should first multiply the first equation by –3 and then add the two equations together to eliminate the x-terms. When he did so, he also eliminated the y-terms and got the equation 0 = 0, so he thought that the system of equations must have an infinite number of solutions. To check his answer, he graphed the equations 3 x minus 2 y = 4 and 9 x minus 6 y = 12 with his graphing calculator, but he could only see one line. Why is this? because the system of equations actually has only one solution because the system of equations actually has no solution because the graphs of the two equations overlap each other because the graph of one of the equations does not exist
Answer: C. Because the graphs of the two equations overlap each other.
Step-by-step explanation: I took the test its right! Hope this helps
A boat is moving due east at 15 miles per hour when it encounters a current LaTeX: \textbf{C} = 2\textbf{i}+17\textbf{j}C = 2 i + 17 j. What is the path of the boat in this current if the boater keeps it pointed due east? What direction should the boater steer in order to go due east?
Answer:
a) [tex]\vec v_{B} - \vec v_{C} = 3\, i - 17\,j[/tex], b) [tex]\theta = 79.992^{\textdegree}[/tex] (clockwise)
Step-by-step explanation:
The resultant velocity of the boat must be:
[tex]\vec v_{B} = 15\,i[/tex]
Likewise, the velocity of the current is:
[tex]\vec v_{C} = 2\,i + 17\,j[/tex]
a) The intended velocity of the boat is:
[tex]\vec v_{B} - \vec v_{C} = (15-2)\,i + (0-17)\,j[/tex]
[tex]\vec v_{B} - \vec v_{C} = 3\, i - 17\,j[/tex]
b) The direction of the boat is:
[tex]\theta = \tan^{-1}\left(\frac{17}{3} \right)[/tex] (clockwise)
[tex]\theta = 79.992^{\textdegree}[/tex] (clockwise)
2. Is the bottled water you’re drinking really purified water? A 4-year study of bottled water brands conducted by the Natural Resources Defense Council found that 25% of bottled water is just tap water packaged in a bottle. Consider a sample of five bottled water brands, and let Y equal the number of these brands that use tap water. A. What is the type of probability distribution for Y. B. Find P(Y=2) C. Find P(Y<1).
Answer:
a.
The random variable has binomial distribution with parameters p = 0.25, n=5.
b.
[tex]P(Y=2) = {5 \choose 2} (0.25)^2(0.75)^3 = 0.264\\P(Y<1 ) = P(Y = 0 ) = (0.75)^5 = 0.237[/tex]
Step-by-step explanation:
a.
Remember what a random variable with binomial distribution is. It is a random variable that counts the number of successes and in n trials of an experiment.
In this case, if you have 0,1,2,3,4,5 bottles of water, your success is that the bottle has tap water, and the probability of that success is p = 0.25. and the number of trials is n = 5.
b.
Using the formula for the binomial distribution you get that
[tex]P(Y=2) = {5 \choose 2} (0.25)^2(0.75)^3 = 0.264\\P(Y<1 ) = P(Y = 0 ) = (0.75)^5 = 0.237[/tex]
A tire company sends an email survey to all customers who purchase new tires each month. The previous month, 201 people purchased new tires. Surveys were sent to 100 of these people, chosen at random, and 51 people responded to the survey. Identify the population and the sample.
Answer:
The population consists of the 201 people who bought tires last month.
The sample consists of the 51 people who responded.
Step-by-step explanation:
In statistical analysis, the population is a set of all values that can be considered for an experiment. It is a set of observations from which a sample is selected. The sample is a subset of the population.
In this case, it is provided that a tire company sends an email survey to all customers who purchase new tires each month.
The number of tires sold last month was N = 201.
The email survey was sent to 100 of these 201 people and 51 responded.
In this case, since the survey is done every month, the population consists of all the people who bought tires the previous month.
So, the population consists of the 201 people who bought tires last month.
The number of people who responded was, n = 51.
The sample consists of these 51 people because the data from these 51 people would be used to conduct and draw conclusion about the survey.
So, the sample consists of the 51 people who responded.