2. The sample data of EU carbon dioxide emissions has a mean of 7.9 and standard deviation of 3.6. (40 points)

If you get both of these questions right I will give you 50 points

Stella [2.4K]

Answer:

1) c 226

2)

1. E

2. D

3. B

4. A

5. C

Step-by-step explanation:

6^3=216

5x(4-2)

5x2=10

216+10=226

8 0

3 years ago

Suppose we want to choose 7 objects, without replacement, from 10 distinct objects.

Citrus2011 [14]

Answer:

= 604800

Step-by-step explanation:

10 permute 7

the order matters

10 x 9 x8 x7 x6 x5 x4

hope this helps

7 0

2 years ago

Find the area<br> cant solve

Alex73 [517]

Answer:

36in²

Step-by-step explanation:

<h3>Area of the White Region: </h3>

A = l * w

The rectangle is 3 by 2.

A = 3 * 2

A = 6

The white part of the rectangle is 6in².

<h3>Area of the blue region:</h3>

A = l * w

The rectangle is 6 by 7.

A = 6*7

A = 42

The blue part of the rectangle is 42in².

<h3>Area of the shaded region:</h3>

[area of the blue part] - [area of the white part]

42 - 6 = 36

The area of the shaded region should be 36in².

8 0

3 years ago

Can someone answer these pls

vazorg [7]

Answer:

x = -2

Step-by-step explanation:

hope this helps you

6 0

3 years ago

Read 2 more answers

A survey collected data on annual credit card charges in seven different categories of expenditures: transportation, groceries,

Andru [333]

Answer:

A. Null and alternative hypothesis:

$H_0: \mu_d=0\\\\H_a:\mu_d\neq 0$

B. Yes. At a significance level of 0.05, there is enough evidence to support the claim that there is signficant difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.

P-value = 0.00002

C. As the difference is calculated as (population 1 − population 2), being population 1: groceries and population 2: dinning out, and knowing there is evidence that the true mean difference is positive, we can say that the groceries annual credit card charge is higher than dinning out annual credit card charge.

The point estimate is the sample mean difference d=$840.

The 95% confidence interval for the mean difference between the population means is (490, 1190).

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that there is signficant difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.

Then, the null and alternative hypothesis are:

$H_0: \mu_d=0\\\\H_a:\mu_d\neq 0$

The significance level is 0.05.

The sample has a size n=42.

The sample mean is M=840.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=1123.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{1123}{\sqrt{42}}=173.2827$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{840-0}{173.2827}=\dfrac{840}{173.2827}=4.848$

The degrees of freedom for this sample size are:

$df=n-1=42-1=41$

This test is a two-tailed test, with 41 degrees of freedom and t=4.848, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=2\cdot P(t>4.848)=0.00002$

As the P-value (0.00002) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

At a significance level of 0.05, there is enough evidence to support the claim that there is signficant difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.

We have to calculate a 95% confidence interval for the mean difference.

The t-value for a 95% confidence interval and 41 degrees of freedom is t=2.02.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_M=2.02 \cdot 173.283=350$

Then, the lower and upper bounds of the confidence interval are:

$LL=M-t \cdot s_M = 840-350=490\\\\UL=M+t \cdot s_M = 840+350=1190$

The 95% confidence interval for the mean difference is (490, 1190).

7 0

3 years ago