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3 years ago

Question 15

Mathematics

1 answer:

Ghella [55]3 years ago

7 0

Answer:

C: -680/9

B: -185/18

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A researcher is interested in determining if the more than two thirds of students would support making the Tuesday before Thanks

Yakvenalex [24]

Answer:

The null hypothesis is, {H₀}: p = 0.66, H₀: p = 0.66 .

The alternative hypothesis is, {Hₐ}: p > 0.66, Hₐ: p > 0.66

Step-by-step explanation:

Let p represent the proportion of students who are supportive of making the day before Thanksgiving a holiday.

Since two populations are considered, the alternative hypothesis for the difference of the means of the two populations has to be stated.

The proportion of two-third students are supportive of making the day before Thanksgiving a holiday is p = 0.66 obtained as shown below:

The null hypothesis is,

{H₀}:p = 0.66, H₀: p = 0.66

The alternative hypothesis is,

{Hₐ:p > 0.66, Ha: p > 0.66

The null hypothesis is, {H₀}: p = 0.66, H₀: p=0.66 .

The alternative hypothesis is, {Hₐ}: p > 0.66, Hₐ: p > 0.66

3 0

3 years ago

if a student has homework 72 days out of 100 what fraction and what percentage of days did the student not have homework?

neonofarm [45]

Answer:

The Student has 72% out of 100%

Step-by-step explanation:

Its the same thing but with fractions

4 0

2 years ago

Could you help me.

Sedbober [7]

Answer:

X-Large

Step-by-step explanation:

8 0

3 years ago

Find the variance of the following data. Round your answer to one decimal place. x 1 2 3 4 5 P(X=x) 0.3 0.2 0.2 0.1 0.2

Reptile [31]

The variance of a distribution is the square of the standard deviation

The variance of the data is 2.2

<h3>How to calculate the variance</h3>

Start by calculating the expected value using:

$E(x) = \sum x* P(x)$

So, we have:

$E(x) = 1 * 0.3 + 2* 0.2 +3 * 0.2 + 4 * 0.1 + 5 * 0.2$

This gives

$E(x) = 2.7$

Next, calculate E(x^2) using:

$E(x^2) = \sum x^2* P(x)$

So, we have:

$E(x^2) = 1^2 * 0.3 + 2^2* 0.2 +3^2 * 0.2 + 4^2 * 0.1 + 5^2 * 0.2$

$E(x^2) = 9.5$

The variance is then calculated as:

$Var(x) = E(x^2) - (E(x))^2$

So, we have:

$Var(x) = 9.5 - 2.7^2$

$Var(x) = 2.21$

Approximate

$Var(x) = 2.2$

Hence, the variance of the data is 2.2