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3 years ago

15

45 girls tried out for the field hockey team. 20% of the girls who tried out did not make the team. How many girls did not make

the team?
It lagged earlier when I first asked

2 answers:

Stella [2.4K]3 years ago

6 0

Answer:

9 girls did not make the team

Step-by-step explanation:

20*45=900

900/100=9

kolbaska11 [484]3 years ago

4 0

Answer:

9 girls didn't make the team

Step-by-step explanation:

The person above did it a great way, and that is the way I use it, but if you want a method that involves fractions here it is: 20% is the same as 1/5. Since there are 45 girls total, we need to find 1/5 of 45 and you can achieve that by multiplying 45 by 1/5 and you get 45/5 which is 9. Hope this helped :) .

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From a random sample of size 18, a researcher states that (11.1, 15.7) inches is a 90% confidence interval for mu, the mean leng

alexandr402 [8]

Complete Question

From a random sample of size 18, a researcher states that (11.1, 15.7) inches is a 90% confidence interval for mu, the mean length of bass caught in a small lake. A normal distribution was assumed. Using the 90% confidence interval obtain:

a. A point estimate of $\mu$ and its 90% margin of error.

b. A 95% confidence interval for $\mu$ .

Answer:

a

$\= x = 13.4$ . $E = 2.3$

b

$10.7 < \mu < 16.1$

Step-by-step explanation:

From the question we are told that

The sample size is n = 18

The 90% confidence interval is (11.1, 15.7)

Generally the point estimate of $\mu$ is mathematically evaluated as

$\= x = \frac{11.1 + 15.7 }{2}$

=> $\= x = 13.4$

Generally the margin of error is mathematically evaluated as

$E = \frac{15.7 - 11.1}{2 }$

=> $E = 2.3$

From the question we are told the confidence level is 90% , hence the level of significance is

$\alpha = (100 - 90 ) \%$

=> $\alpha = 0.10$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.645$

Generally the equation for the lower limit of the confidence interval is

$\= x - Z_{\frac{\alpha }{2} } * \frac{s}{\sqrt{18} } = 11.1$

=> $13.4 - 0.3877 s = 11.1$

=> $s = 5.932$

From the question we are told the confidence level is 95% , hence the level of significance is

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

=> $\alpha = 0.05$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

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

Generally the margin of error is mathematically represented as

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

=> $E = 1.96 * \frac{5.932}{\sqrt{18} }$

=> $E = 2.7$

Generally 95% confidence interval is mathematically represented as

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

=> $13.4 - 2.7 < \mu < 13.4 + 2.7$

=> $10.7 < \mu < 16.1$

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3 years ago