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4 years ago

14

How do I Invert 6/7? Can you please help me!

1 answer:

Dmitry [639]4 years ago

3 0

Answer:

<h2>0. 857142</h2>

i hope this helps

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Find n. Perimeter = 86 in.

FromTheMoon [43]

Let the length of rectangle be L and the width of rectangle be W.
Since length exceeds the width by 25 inches, length will be
L = W + 25

Now the perimeter, P, is given by
P = 2(L + W)
Substituting L = W + 25 in the above equation,
P = 2(W + 25 + W)
P = 2(2W + 25)
P = 4W + 50
But P = 86 inches
P = 4W + 50 = 86
4W = 86 - 50 = 36
W = 36/4 = 9

Hence, width W = 9 inches.
Length L = W + 25 = 9 + 25 = 34 inches.

5 0

4 years ago

The product of 11 and 5 times a number mutplied by 4

Vlada [557]

11 + 5 x 4a

Hope this helps! Do you need an explanation?

3 0

3 years ago

Mr. Hills horses, all together, eat 395 pounds of oats each month. If Mr. Hills has 5 horses, and each horse eats the sane amoun

Marizza181 [45]

I would say 80, because without an estimate it'd be 79 and 79x5=395. So 80x5=400, which is the estimate of 395. So 80 is the answer

8 0

4 years ago

Fill in blanks to write the function when x=3 h(x)=-4-7, when x=3:<br><br>h(__)=__

Veseljchak [2.6K]

Answer:

h(3) = -3

Step-by-step explanation:

h(3) = 4-7

h(3) = -3

6 0

3 years ago

Read 2 more answers

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: $15.3\pm0.198$

OR

(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