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3 years ago

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3 5. The cost of a reserved seat is 1 3/4 times the cost of general admission. If a reserved seat is $14, what is the price of g

eneral admission.

1 answer:

SIZIF [17.4K]3 years ago

3 0

$14 / 1.75 = $8

General admission is $8

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26. Students who take a statistics course are given a pre-test on the concepts and skills for the first chapter of a statistics

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The test statistic value lies to the right of the critical value. So we have sufficient evidence to reject the null hypothesis.

<h3>What are null hypotheses and alternative hypotheses?</h3>

In null hypotheses, there is no relationship between the two phenomenons under the assumption or it is not associated with the group. And in alternative hypotheses, there is a relationship between the two chosen unknowns.

Students who take a statistics course are given a pre-test on the concepts and skills for the first chapter of a statistics course.

Then they are given a post-test once the professor has concluded lecturing on the material.

Pre-test and post-test scores for 4 students in an elementary statistics class are given below.

Then we have

$\mu _d = \mu _{post} - \mu _{pre}$

Then the null hypotheses and alternative hypotheses will be

H₀: $\mu _d = 0$

Hₐ: $\mu _d > 0$

Then the test statistic will be

$\rm \overline{x} _d = \dfrac{\Sigma x_d}{n} = \dfrac{15+12+10+1}{4}\\\\\overline{x} _d = 9.5$

Then

$\rm S_d = 6.02$

The test statistic value is given by

$\rm t = \dfrac{\overline{x} _d }{\dfrac{S_d}{\sqrtn}} \\\\t = \dfrac{9.5}{\dfrac{6.02}{\sqrt4}}\\\\t = 3.16$

Since this is a right-tailed test, so the critical value is given by

$\rm t_{n-1}(\alpha ) = t_3 (0.05) = 2.353$

Since the test statistic value lies to the right of the critical value. So we have sufficient evidence to reject the null hypothesis.

Hence, we can conclude that $\mu _d > 0$ that is test scores have improved.

More about the null hypotheses and alternative hypotheses link is given below.

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2 years ago

What is the base 10 representation of 1110 base 2?

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An art history professor assigns letter grades on a test according to the following scheme.A: Top 5% of scoresB: Scores below th

adoni [48]

Answer:

Grade B score:

$76 \leq x \leq 89$

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 73.3

Standard Deviation, σ = 9.7

We are given that the distribution of score on test is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

B: Scores below the top 5% and above the bottom 62%

We have to find the value of x such that the probability is 0.62

$P( X > x) = P( z > \displaystyle\frac{x - 73.3}{9.7})=0.62$

$= 1 -P( z \leq \displaystyle\frac{x - 73.3}{9.7})=0.62$

$=P( z \leq \displaystyle\frac{x - 73.3}{9.7})=0.38$

Calculation the value from standard normal z table, we have,

$\displaystyle\frac{x - 73.3}{9.7} = 0.305\\\\x = 76.26$

We have to find the value of x such that the probability is 0.05

$P(X < 0.95) = \\\\P( X < x) = P( z < \displaystyle\frac{x - 73.3}{9.7})=0.95$

Calculation the value from standard normal z table, we have,

$\displaystyle\frac{x - 73.3}{9.7} = 1.645\\\\x = 89.26$

Thus, the numerical value of score to achieve grade B is

$76 \leq x \leq 89$

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