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exis [7]
3 years ago
7

Solve for the indicated variable.  for r

Mathematics
1 answer:
Sergeeva-Olga [200]3 years ago
8 0
What is the indicated variable
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Gas prices have really decreased in the past month. At the beginning of March I paid $2.15 per gallon. I just recently filled my
s344n2d4d5 [400]

Answer: 27.44 % decrease in gas prices

Step-by-step explanation: First subtract 2.15 by 1.56 to get 0.59

then divide it by the starting value which is 2.15

0.59 divided by 2.15 is  approximately 0.2744

then multiply times 100 to get 27.44 %

hope this helps mark me brainliest if it helped

7 0
3 years ago
A marketing researcher wants to test the hypothesis that, within the population, there are differences in the importance attache
Ira Lisetskai [31]

Answer: Analysis of variance

Step-by-step explanation:

Analysis of variance is the statistical test that's used in analyzing the differences among means. The analysis of variance is used to determine if a statistically significant differences exust between the means of the independent groups.

Based on the question given, the null hypothesis will be that no difference in the importance that's attached to shopping by the consumers living in different regions in the United States.

6 0
3 years ago
3. Tyler and his family went on a backpacking trip. They started hiking at 3:00 P.M. It took 2 hours and 30 minutes to reach the
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I'm pretty sure the answer is 7:00 pm
4 0
3 years ago
Whats the radius of d = 29 m
VMariaS [17]

Answer:

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5 0
3 years ago
Read 2 more answers
The following results come from two independent random samples taken of two populations.
photoshop1234 [79]

Answer:

(a)\ \bar x_1 - \bar x_2 = 2.0

(b)\ CI =(1.0542,2.9458)

(c)\ CI = (0.8730,2.1270)

Step-by-step explanation:

Given

n_1 = 60     n_2 = 35      

\bar x_1 = 13.6    \bar x_2 = 11.6    

\sigma_1 = 2.1     \sigma_2 = 3

Solving (a): Point estimate of difference of mean

This is calculated as: \bar x_1 - \bar x_2

\bar x_1 - \bar x_2 = 13.6 - 11.6

\bar x_1 - \bar x_2 = 2.0

Solving (b): 90% confidence interval

We have:

c = 90\%

c = 0.90

Confidence level is: 1 - \alpha

1 - \alpha = c

1 - \alpha = 0.90

\alpha = 0.10

Calculate z_{\alpha/2}

z_{\alpha/2} = z_{0.10/2}

z_{\alpha/2} = z_{0.05}

The z score is:

z_{\alpha/2} = z_{0.05} =1.645

The endpoints of the confidence level is:

(\bar x_1 - \bar x_2) \± z_{\alpha/2} * \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}

2.0 \± 1.645 * \sqrt{\frac{2.1^2}{60}+\frac{3^2}{35}}

2.0 \± 1.645 * \sqrt{\frac{4.41}{60}+\frac{9}{35}}

2.0 \± 1.645 * \sqrt{0.0735+0.2571}

2.0 \± 1.645 * \sqrt{0.3306}

2.0 \± 0.9458

Split

(2.0 - 0.9458) \to (2.0 + 0.9458)

(1.0542) \to (2.9458)

Hence, the 90% confidence interval is:

CI =(1.0542,2.9458)

Solving (c): 95% confidence interval

We have:

c = 95\%

c = 0.95

Confidence level is: 1 - \alpha

1 - \alpha = c

1 - \alpha = 0.95

\alpha = 0.05

Calculate z_{\alpha/2}

z_{\alpha/2} = z_{0.05/2}

z_{\alpha/2} = z_{0.025}

The z score is:

z_{\alpha/2} = z_{0.025} =1.96

The endpoints of the confidence level is:

(\bar x_1 - \bar x_2) \± z_{\alpha/2} * \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}

2.0 \± 1.96 * \sqrt{\frac{2.1^2}{60}+\frac{3^2}{35}}

2.0 \± 1.96* \sqrt{\frac{4.41}{60}+\frac{9}{35}}

2.0 \± 1.96 * \sqrt{0.0735+0.2571}

2.0 \± 1.96* \sqrt{0.3306}

2.0 \± 1.1270

Split

(2.0 - 1.1270) \to (2.0 + 1.1270)

(0.8730) \to (2.1270)

Hence, the 95% confidence interval is:

CI = (0.8730,2.1270)

8 0
3 years ago
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