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3 years ago

Find f(-1) please help, not good at math.

Mathematics

1 answer:

bagirrra123 [75]3 years ago

7 0

This means that X in the equation equals -1

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0.941176471 to the nearest hundredth ?

Wewaii [24]

0.941176471 to the nearest hundredth is 0.94

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3 years ago

Which category does each card go into?

WINSTONCH [101]

Step-by-step explanation:

E3, E6, E11, E7: no solution

E1, E12 : 1 solution

E2, E8, E10 : infinite solutions

7 0

3 years ago

Read 2 more answers

Write 79/10 as a mixed number

oee [108]

Answer:

7 9/10

Step-by-step explanation:

cuz mah brain big

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3 years ago

What is the difference between the areas of a circular table with diameter 6 ft and a circular table with diameter 8 ft? Round y

MrMuchimi

If the diameter is given in a problem, area is computed as

A = (π/4) × D2 where π is equal to 3.14

If D (diameter) of one circular table is 6 ft its Area is 28 sq ft.

If D of another circular is 8 ft its Area is 50 sq ft.

The difference between the areas of the two tables is, therefore (50-28) or 22 sq. ft.

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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)

5 0

3 years ago