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3 years ago

The smallest square number which can be divisible by 11 and 5

Mathematics

1 answer:

Elena L [17]3 years ago

6 0

Answer:

3025

Step-by-step explanation:

let's look at it in this concept.

For a number to be divisible by 11 and 5. it must be a multiple of the LCM of 11 and 5.

LCM of 11 and 5=55

therefore the number is 55x, where x is a positive integer.

it is a said that the number is a perfect square

therefore the square root of 55x must be an integer.

$\sqrt{55x} = \sqrt{5 \times 11 \times x}$

the smallest value of x to make 55x a perfect square is....

$x = 11 \times 5 = 55$

Therefore the number is.... .

$55x = 55(55) = 3025$

sweet right

<h2>BRAINLIEST PLS....</h2>

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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.