HELP!!! FOR TEN POINTS

Find the common difference of the arithmetic sequence.

slavikrds [6]

Each time you are subtracting 1/9 to get the next term

For example, going from term 2 to term 3 is a difference of -1/9

(1 8/9) - (1/9) = 1 7/9

which is why the answer is choice A

7 0

2 years ago

If a box is 12.2 cm by 9.8 cm by 16.4 cm, what is the volume of the box?

sp2606 [1]

To find volume multiply length, width, and height.
12.2 x 9.8 x 16.4 =1960.784
The answer is 1960.748 cm.

4 0

3 years ago

An air transport association surveys business travelers to develop quality ratings for transatlantic gateway airports. The maxim

Flura [38]

Answer:

The 95% confidence interval estimate of the population mean rating for Miami is (6.0, 7.5).

Step-by-step explanation:

The (1 - α)% confidence interval for the population mean, when the population standard deviation is not provided is:

$CI=\bar x \pm t_{\alpha/2, (n-1)}\cdot \frac{s}{\sqrt{n}}$

The sample selected is of size, n = 50.

The critical value of t for 95% confidence level and (n - 1) = 49 degrees of freedom is:

$t_{\alpha/2, (n-1)}=t_{0.05/2, 49}\approx t_{0.025, 60}=2.000$

*Use a t-table.

Compute the sample mean and sample standard deviation as follows:

$\bar x=\frac{1}{n}\sum X=\frac{1}{50}\times [1+5+6+...+10]=6.76\\\\s=\sqrt{\frac{1}{n-1}\sum (x-\bar x)^{2}}=\sqrt{\frac{1}{49}\times 31.12}=2.552$

Compute the 95% confidence interval estimate of the population mean rating for Miami as follows:

$CI=\bar x \pm t_{\alpha/2, (n-1)}\cdot \frac{s}{\sqrt{n}}$

$=6.76\pm 2.000\times \frac{2.552}{\sqrt{50}}\\\\=6.76\pm 0.722\\\\=(6.038, 7.482)\\\\\approx (6.0, 7.5)$

Thus, the 95% confidence interval estimate of the population mean rating for Miami is (6.0, 7.5).

7 0

2 years ago

A consumer group has determined that the distribution of life spans for gas ranges (stoves) has a mean of 15.0 years and a stand

Art [367]

Answer:

b. Mean = 1.6 years, standard deviation - 0.92 years, shape: approximately Normal.

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Subtraction of normal Variables:

When we subtract normal variables, the mean is the subtraction of the means, while the standard deviation is the square root of the sum of the variances.

A consumer group has determined that the distribution of life spans for gas ranges (stoves) has a mean of 15.0 years and a standard deviation of 4.2 years. Sample of 35:

This means that:

$\mu_G = 15$

$s_G = \frac{4.2}{\sqrt{35}} = 0.71$

The distribution of life spans for electric ranges has a mean of 13.4 years and a standard deviation of 3.7 years. Sample of 40:

This means that:

$\mu_E = 13.4$

$s_E = \frac{3.7}{\sqrt{40}} = 0.585$

Which of the following best describes the sampling distribution of the difference in mean life span of gas ranges and electric ranges?

Shape is approximately normal.

Mean:

$\mu = \mu_G - \mu_E = 15 - 13.4 = 1.6$

Standard deviation:

$s = \sqrt{s_G^2+s_E^2} = \sqrt{0.71^2+0.585^2} = 0.92$

So the correct answer is given by option b.

8 0

3 years ago

According to the Vivino website, suppose the mean price for a bottle of red wine that scores 4.0 or higher on the Vivino Rating

Natali [406]

Answer:

a) Null and alternative hypothesis

$H_0: \mu=32.48\\\\H_a:\mu< 32.48$

b) Test statistic t=-1.565

P-value = 0.0612

NOTE: the sample size is n=65.

c) Do not reject H0. We cannot conclude that the price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

d) Null and alternative hypothesis

$H_0: \mu=32.48\\\\H_a:\mu< 32.48$

Test statistic t=-1.565

Critical value tc=-1.669

t>tc --> Do not reject H0

Do not reject H0. We cannot conclude that the price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the mean price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

Then, the null and alternative hypothesis are:

$H_0: \mu=32.48\\\\H_a:\mu< 32.48$

The significance level is 0.05.

The sample has a size n=65.

The sample mean is M=30.15.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=12.

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

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{12}{\sqrt{65}}=1.4884$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{30.15-32.48}{1.4884}=\dfrac{-2.33}{1.4884}=-1.565$

The degrees of freedom for this sample size are:

$df=n-1=65-1=64$

This test is a left-tailed test, with 64 degrees of freedom and t=-1.565, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=P(t$

As the P-value (0.0612) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

At a significance level of 0.05, there is not enough evidence to support the claim that the mean price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

Critical value approach

At a significance level of 0.05, for a left-tailed test, with 64 degrees of freedom, the critical value is t=-1.669.

As the test statistic is greater than the critical value, it falls in the acceptance region.

The null hypothesis failed to be rejected.

6 0

2 years ago