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3 years ago

6

If the following object is translated T(-1, 4), where will the translation be located? I have attached the rest of the question

in images.

1 answer:

Juliette [100K]3 years ago

8 0

The fourth attachment is the correct answer.

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A college requires applicants to have an ACT score in the top 12% of all test scores. The ACT scores are normally distributed, w

DochEvi [55]

Answer:

a) The lowest test score that a student could get and still meet the colleges requirement is 27.0225.

b) 156 would be expected to have a test score that would meet the colleges requirement

c) The lowest score that would meet the colleges requirement would be decreased to 26.388.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 21.5, \sigma = 4.7$

a. Find the lowest test score that a student could get and still meet the colleges requirement.

This is the value of X when Z has a pvalue of 1 - 0.12 = 0.88. So it is X when Z = 1.175.

$Z = \frac{X - \mu}{\sigma}$

$1.175 = \frac{X - 21.5}{4.7}$

$X - 21.5 = 1.175*4.7$

$X = 27.0225$

The lowest test score that a student could get and still meet the colleges requirement is 27.0225.

b. If 1300 students are randomly selected, how many would be expected to have a test score that would meet the colleges requirement?

Top 12%, so 12% of them.

0.12*1300 = 156

156 would be expected to have a test score that would meet the colleges requirement

c. How does the answer to part (a) change if the college decided to accept the top 15% of all test scores?

It would decrease to the value of X when Z has a pvalue of 1-0.15 = 0.85. So X when Z = 1.04.

$Z = \frac{X - \mu}{\sigma}$

$1.04 = \frac{X - 21.5}{4.7}$

$X - 21.5 = 1.04*4.7$

$X = 26.388$

The lowest score that would meet the colleges requirement would be decreased to 26.388.

6 0

4 years ago

The owner of Mattâ€™s Gas Station wants to study gasoline purchases of customers at his station. In 2017, the mean amount of gas

gulaghasi [49]

Answer:

Step-by-step explanation:

a) We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

µ = 9

For the alternative hypothesis,

µ ≠ 9

This is a 2 tailed test

Since the population standard deviation is given, z score would be determined from the normal distribution table. The formula is

z = (x - µ)/(σ/√n)

Where

x = sample mean

µ = population mean

σ = population standard deviation

n = number of samples

From the information given,

µ = 9

x = 9.9

σ = 2.5

n = 50

z = (9.9 - 9)/(2.5/√50) = 2.55

Looking at the normal distribution table, the probability corresponding to the area above the z score is 1 - 0.99461 = 0.00539

Recall, population mean is 9

The difference between sample sample mean and population mean is 9.9 - 9 = 0.9

Since the curve is symmetrical and it is a two tailed test, the x value for the left tail is 9 - 0.9 = 8.1

the x value for the right tail is 9 + 0.9 = 9.9

These means are higher and lower than the null mean. Thus, they are evidence in favour of the alternative hypothesis. We will look at the area in both tails. Since it is showing in one tail only, we would double the area. We already got the area above the z score as z = 0.00539

We would double this area to include the area in the left tail of z = - 2.55. Thus

p = 0.00539 × 2 = 0.01078

Since alpha, 0.01 < 0.01078, we would reject the null hypothesis

But we are told to use the critical value approach, then

Since α = 0.01, the critical value is determined from the normal distribution table.

For the left, α/2 = 0.01/2 = 0.005

The z score for an area to the left of 0.005 is - 2.575

For the right, α/2 = 1 - 0.005 = 0.995

The z score for an area to the right of 0.995 is 2.575

In order to reject the null hypothesis, the test statistic must be smaller than - 2.575 or greater than 2.575

Since - 2.55 > - 2.575 and 2.55 < 2.575, we would reject the null hypothesis. This corresponds to our previous decision.

b) we would assume normal distribution because the sample size is sufficiently large and the population standard deviation is known.

c) Confidence interval is written in the form,

(Sample mean ± margin of error)

Since the sample size is large and the population standard deviation is known, we would use the following formula and determine the z score from the normal distribution table.

Margin of error = z × σ/√n

The z score for confidence level of 99% is 2.58

Margin of error = 2.58 × 2.5/√50 = 0.91

Confidence interval = 9.9 ± 0.91

The lower end of the confidence interval is

9.9 - 0.91 = 8.99

Therefore, p = 9 will be contained in the interval.

4 0

3 years ago

Help meeeee!!!!!! Please

Over [174]

Answer:

-1/11

Step-by-step explanation:

if this is one of the answer choices.

6 0

3 years ago

Permutations: 5!·4! is equivalent to what (5!)(4!)^2, (5)(4!)^2 or 20! ?

solniwko [45]

Answer:

5!*4! is equivalent to 5*(4!)^2

Step-by-step explanation:

5!*4! = 2880

5*(4!)^2 = 2880

4 0

3 years ago

The steps below show the incomplete solution to find the value of m for the equation 4m − 2m + 4 = −1 + 13:

Law Incorporation [45]

The second 2m=8
2m=12-4

7 0

3 years ago

Read 2 more answers