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3 years ago

Of the pets in the pet show,

Mathematics

2 answers:

Arada [10]3 years ago

5 0

Answer:

1 17/40

Step-by-step explanation:

$\frac{5}{8} + \frac{4}{5} \\$

Find the common denominator, which is 40

so then 5/8 = 25/40 and 4/5 = 32/40

$\frac{25}{40} + \frac{32}{40} = \frac{57}{40} = 1 \frac{17}{40}$

Kay [80]3 years ago

4 0

Answer:

45/58 it is already in simplest form

Step-by-step explanation:

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Angle PSR measures 99°.

Maru [420]

Answer:

Step-by-step explanation:

7 0

3 years ago

What precent of 95 is 60.8

dezoksy [38]

60.8 is 64% of 95.

Step-by-step explanation:

Given number is;

We have to find at what percent, the value is 60.8

Let x be the percent.

Therefore;

x% of 95 = 60.8

$\frac{95x}{100}=60.8 \\0.95x=60.8$

Dividing both sides by 0.95

$\frac{0.95x}{0.95}=\frac{60.8}{0.95}\\x=64$

60.8 is 64% of 95.

Keywords: percent, division

Learn more about division at:

#LearnwithBrainly

5 0

3 years ago

What is 8/10 simplified by

Oksanka [162]

Answer:

4/5

Step-by-step explanation:

that is it

6 0

2 years ago

Read 2 more answers

Experian would like to test the hypothesis that the average credit score for an adult in Virginia is different from the average

aliya0001 [1]

Answer:

a. We fail to reject the hypothesis that the average credit score for an adult in Virginia is different from the average credit score for an adult in North Carolina at the significance level of 0.05.

b. The 95% confidence interval for the true difference of means is -2.2468 and 36.2468. There is a probability of 95% that the true difference of means $\mu_{1}-\mu_{2}$ is between -2.2468 and 36.2468. This confidence interval contains the number 0, this is consistent with the results we got in a. Because we fail to reject the hypothesis that the average credit score for an adult in Virginia is different from the average credit score for an adult in North Carolina at the significance level of 0.05.

Step-by-step explanation:

Let $\mu_{1}-\mu_{2}$ be the true difference between the average credit score for an adult in Virginia and the average credit score for an adult in North Carolina. We have the large sample sizes $n_{1} = 40$ and $n_{2} = 35$ , the unbiased point estimate for $\mu_{1}-\mu_{2}$ is $\bar{x}_{1} - \bar{x}_{2}$ , i.e., 699-682 = 17.

The standard error is given by $\sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}$ , i.e.,

$\sqrt{\frac{(44)^{2}}{40}+\frac{(41)^{2}}{35}}$ = 9.8198.

a. We want to test $H_{0}: \mu_{1}-\mu_{2} = 0$ vs $H_{1}: \mu_{1}-\mu_{2} \neq 0$ (two-tailed alternative). The rejection region is given by RR = {z | z < -1.96 or z > 1.96} where -1.96 and 1.96 are the 2.5th and 97.5th quantiles of the standard normal distribution respectively. The test statistic is $Z = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{\sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}}$ and the observed value is $z_{0} = \frac{17}{9.8198} = 1.7312$ . Because 1.7312 does not fall inside RR, we fail to reject the null hypothesis.

b. The endpoints for a 95% confidence interval for $\mu_{1}-\mu_{2}$ is given by $17\pm (z_{0.05/2})9.8198$ , i.e., $17\pm (z_{0.025})9.8198$ where $z_{0.025}$ is the 2.5th quantile of the standard normal distribution, i.e., -1.96, so, we have 17-(1.96)(9.8198) and 17+(1.96)(9.8198), i.e., -2.2468 and 36.2468. There is a probability of 95% that the true difference of means $\mu_{1}-\mu_{2}$ is between -2.2468 and 36.2468. This confidence interval contains the number 0, this is consistent with the results we got in a. Because we fail to reject the hypothesis that the average credit score for an adult in Virginia is different from the average credit score for an adult in North Carolina at the significance level of 0.05.

3 0

3 years ago

5x + 8 − 3x = −10 x = −9 x = −1 x = 1 x = 9

zvonat [6]

Answer:

-9

Step-by-step explanation:

7 0

3 years ago