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3 years ago

I need help with my super haaaaaaaaaard work

Mathematics

2 answers:

Flauer [41]3 years ago

7 0

Answer:

What is the Question so I can answer?

Step-by-step explanation:

Nat2105 [25]3 years ago

6 0

Answer:

where's the "super hard work"???????????

Step-by-step explanation:

There's no work, I can't help you if you don't show what you need help with.

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3 years ago

Somebody plz help me

nata0808 [166]

Answer:

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2 years ago

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3 years ago

Point T is on a line with a slope of -2. Which of these could be the coordinates of

Furkat [3]

Answer: A.(3,1)

D.(2,7) both are correct

Step-by-step explanation:

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3 years ago

A random sample of 250 students at a university finds that these students take a mean of 15.3 credit hours per quarter with a st

SpyIntel [72]

Answer: $15.3\pm0.198$

(15.102, 15.498)

Step-by-step explanation:

The formula to find the confidence interval $(\mu)$ is given by :-

$\overline{x}\pm z_{\alpha/2, df}\dfrac{s}{\sqrt{n}}$

, where n is the sample size

$s$ = sample standard deviation.

$\overline{x}$ = Sample mean

$z_{\alpha/2}$ = Two tailed z-value for significance level of $\alpha$ .

Given : Confidence level = 95% = 0.95

Significance level = $\alpha=1-0.95=0.05$

$s= 1.6$

$\overline{x}=15.3$

sample size : n= 250 , which is extremely large ( than n=30) .

So we assume sample standard deviation is the population standard deviation.

thus , $\sigma=1.6$

By standard normal distribution table ,

Two tailed z-value for Significance level of 0.05 :

$z_{\alpha/2}=z_{0.025}=1.96$

Then, the 95% confidence interval for the mean credit hours taken by a student each quarter :-

$15.3\pm (1.96)\dfrac{1.6}{\sqrt{250}}\\\\ =15.3\pm 0.19833\\\\=\approx15.3\pm0.198\\\\=(15.3-0.198,\ 15.3+0.198)=(15.102,\ 15.498)$

Hence, the mean credit hours taken by a student each quarter using a 95% confidence interval. =(15.102, 15.498)

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3 years ago