What fraction of the original sample remains ?

How many times smaller is 2 × 10-3 than 4 × 10-2? PLEASE HELP

NNADVOKAT [17]

<h2>The required "option A) 20" is correct.</h2>

Step-by-step explanation:

Let x be the smaller than 4 × $10^{-2}$ .

To find, the number of times smaller is 2 × $10^{-3}$ than 4 × $10^{-2}$ = ?

∴ x = $\dfrac{4\times 10^{-2}}{2\times 10^{-3}}$

= 2 × $10^{-2}$ × $10^{3}$

Using the identity,

$a^{m}=\dfrac{1}{a^{-m}}$

= 2 × $10^{-2+3}$

Using the identity,

$a^{m} \timesa^{n}=a^{m+n}$

= 2 × $10^{1}$

= 2 × 10

= 20

Thus, the required "option A) 20" is correct.

6 0

3 years ago

A principal of $3700 is invested at 4.75% interest, compounded annually. How much will the investment be worth after 9 years?

kipiarov [429]

Amount = 3,700 * (1.0475)^9
Amount = 3,700 * 1.5184003135 
Amount = 5,618.08

7 0

3 years ago

Write and solve an equation based on the following question.

Ket [755]

Answer:

D

Step-by-step explanation:

hihihihihuiuiijihihihihihihih

5 0

2 years ago

Can you please explain how to get the answer. I don't understand. I will mark you brainliest!!

larisa86 [58]

Answer: option a its 2

Step-by-step explanation:

6 0

3 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

4 years ago