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3 years ago

What percent of the first 20 natural numbers are prime numbers?

Mathematics

2 answers:

BARSIC [14]3 years ago

6 0

Answer: The required percentage is 40%.

Step-by-step explanation: We are given to find the percentage of the first 20 natural numbers that are prime.

We know that

first 20 natural numbers are 1, 2, 3, 4, . . . , 19, 20.

And, the prime numbers among them are 2, 3, 5, 7, 11, 13, 17 and 19.

So, if S be the sample space for the experiment and A be the event of selecting a prime number.

Then, we have

n(S) = 20 and n(A) = 8.

Therefore, the probability of event A is given by

$P(A)=\dfrac{n(A)}{n(S)}=\dfrac{8}{20}=\dfrac{2}{5}\times100\%=40\%.$

Thus, the required percentage is 40%.

Brilliant_brown [7]3 years ago

3 0

The primes are 2, 3, 5, 7, 11, 13, 17, 19. There are 8 of them, so they make up 40% of the numbers 1–20.

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Is there statistically significant evidence that the districts with smaller classes have higher average test scores? The t-sta

Musya8 [376]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The 95% confidence interval is $[670.03 , 673.97 ]$

The test statistics is $t = 7.7$

The p-value is $p-value = 0$

The p-value suggests that the null hypothesis is rejected with a high degree of confidence. Hence there is statistically significant evidence that the districts with smaller classes have higher average test score

Step-by-step explanation:

From the question we are told that

The sample size is n = 408

The sample mean is $\= y = 672.0$

The standard deviation is $s = 20.3$

Given that the confidence level is 95% then the level of significance is

$\alpha = (100 -95 )\% = 0.05$

From the normal distribution table the critical value of $\frac{\alpha }{2} = \frac{0.05 }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{s}{\sqrt{n} }$

=> $E = 1.96 * \frac{20.3}{\sqrt{408} }$

=> $E = 1.970$

Generally the 95% confidence interval is mathematically represented as

$\= y -E < \mu < \= y + E$

=> $672.0 -1.970 < \mu < 672.0 +1.970$

=> $670.03 < \mu < 673.97$

=> $[670.03 , 673.97 ]$

From the question we are told that

Class size small large

$Avg.score(\= y)$ $\= y_1 = 683.7$ $\= y_2 = 676.0$

$S_y$ $S_{y_1} =20.2$ $S_{y_2} = 18.6$

sample size $n_1 = 229$ $n_2 = 184$

The null hypothesis is $H_o : \mu_1 - \mu_2 = 0$

The alternative hypothesis is $H_a : \mu_1 - \mu_2 > 0$

Generally the standard error for the difference in mean is mathematically represented as

$SE = \sqrt{\frac{S_{y_1}^2 }{n_1} +\frac{S_{y_2}^2 }{n_2} }$

=> $SE = \sqrt{20.2^2 }{229} +\frac{18.6^2 }{184_2} }$

=> $SE = 1.913$

Generally the test statistics is mathematically represented as

$t = \frac{\= y _1 - \= y_2 }{SE}$

=> $t = \frac{683.7 - 676.0 }{1.913}$

=> $t = 7.7$

Generally the p-value is mathematically represented as

$p-value = P(t > 7.7 )$

From the z-table

$P(t > 7.7 ) = 0$

$p-value = 0$

From the values we obtained and calculated we can see that $p-value < \alpha$

This mean that

The p-value suggests that the null hypothesis is rejected with a high degree of confidence. Hence there is statistically significant evidence that the districts with smaller classes have higher average test score