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Nata [24]
3 years ago
14

Solve the equation or formula for the indicated variable a=4b^5h

Mathematics
1 answer:
melisa1 [442]3 years ago
5 0
Given that the formula we are required to solve for h:
a=4b^5h
to solve for h we divide both sides by 4, this will give us:
a/4=(4b^5h)/4
a/4=b^5h
next we divide both sides by b^5, this will give us:
a/(4×b^5)=(b^5h)/b^5
a/(4b^5)=h
hence the answer is:
h=a/(4b^5)



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Answer:

C)265

Step-by-step explanation:

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In this case there are two middle number so you have to add them together then devide them by 2.

700, 740, <u>810, 900</u>, 1050, 1120.

810 + 900 = 1,710

1,710/2 = 855.

Your median is 855.

To find the difference you just subtract the two numbers.

1,120-855= 265.

7 0
3 years ago
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4 years ago
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Anna71 [15]
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6 0
3 years ago
From a random sample of 58 businesses, it is found that the mean time the owner spends on administrative issues each week is 21.
zzz [600]

Answer: (20.86, 22.52)

Step-by-step explanation:

Formula to find the confidence interval for population mean :-

\overline{x}\pm z^*\dfrac{\sigma}{\sqrt{n}}

, where \overline{x} = sample mean.

z*= critical z-value

n= sample size.

\sigma = Population standard deviation.

By considering the given question , we have

\overline{x}= 21.69

\sigma=3.23

n= 58

Using z-table, the critical z-value for 95% confidence = z* = 1.96

Then, 95% confidence interval for the amount of time spent on administrative issues will be :

21.69\pm (1.96)\dfrac{3.23}{\sqrt{58}}

=21.69\pm (1.96)\dfrac{1.7}{7.61577}

=21.69\pm (1.96)(0.223221)

\approx21.69\pm0.83

=(21.69-0.83,\ 21.69+0.83)=(20.86,\ 22.52)

Hence, the 95% confidence interval for the amount of time spent on administrative issues = (20.86, 22.52)

7 0
3 years ago
Steve order plaster cases for storing his baseball cards. Each case has a length of 12 centmeters a width of 6.5 centmeters and
HACTEHA [7]
The volume of the case =
length x width x height
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3 years ago
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