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natita [175]
3 years ago
7

Simplify n^4 divided by n1/2

Mathematics
1 answer:
agasfer [191]3 years ago
5 0

Given:

Consider the expression is

\dfrac{n^4}{n^{\frac{1}{2}}}

To find:

The simplified form of the given expression.

Solution:

We have,

\dfrac{n^4}{n^{\frac{1}{2}}}

Using the properties of exponents, we get

=n^{4-\frac{1}{2}}               \left[\because \dfrac{a^m}{a^n}=a^{m-n}\right]

=n^{\frac{8-1}{2}}

=n^{\frac{7}{2}}

Therefore, the simplified form of the given expression is n^{\frac{7}{2}}.

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How many gallons do you have if you have 18 quarts
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Answer:

4.5 gallons.

Step-by-step explanation:

Multiply the volume value by 4.

In this case, you multiple 4.5 by 4 which equals 18.

4.5 gallons = 18 quarts

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Three times the sum of 4 and a number is twice the<br> number. Find the number.
zavuch27 [327]

Step-by-step explanation:

let number be x

3(4+x)=2x

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12 = 2x - 3x

12 = -x

x= -12

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3 years ago
Use the graph to describe the rule, write the ordered pairs of the image.
maks197457 [2]

Step-by-step explanation:

Both are translations,

1.) Translation of (5,7)

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3 0
2 years ago
Joe collected 94 stamps altogether. He collected 27 in April and 34 in May. How many did he collect in June?
stepan [7]

Answer:

33

Step-by-step explanation:

From what we know in April and May combined Joe has collected 61 stamps (27+34=61). To then work out how many Joe collected in June you would take the 61 we know he has already got away from the grand total of 94 (94-61=33) this means in June he collects 33 stamps. Hope this helps :)

8 0
2 years ago
2. When a large truckload of mangoes arrives at a packing plant, a random sample of 150 is selected and examined for
kirza4 [7]

a) The 90% confidence interval of the percentage of all mangoes on the truck that fail to meet the standards is: (7.55%, 12.45%).

b) The margin of error is: 2.45%.

c) The 90% confidence is the level of confidence that the true population percentage is in the interval.

d) The needed sample size is: 271.

<h3>What is a confidence interval of proportions?</h3>

A confidence interval of proportions has the bounds given by the rule presented as follows:

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

The margin of error is given by:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

The variables are listed as follows:

  • \pi is the sample proportion, which is also the estimate of the parameter.
  • z is the critical value.
  • n is the sample size.

The confidence level is of 90%, hence the critical value z is the value of Z that has a p-value of \frac{1+0.90}{2} = 0.95, so the critical value is z = 1.645.

The sample size and the estimate are given as follows:

n = 150, \pi = \frac{15}{150} = 0.1

The margin of error is of:

M = z\sqrt{\frac{0.1(0.9)}{150}} = 0.0245 = 2.45\%

The interval is given by the estimate plus/minus the margin of error, hence:

  • The lower bound is: 10 - 2.45 = 7.55%.
  • The upper bound is: 10 + 2.45 = 12.45%.

For a margin of error of 3% = 0.03, the needed sample size is obtained as follows:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.03 = 1.645\sqrt{\frac{0.1(0.9)}{n}}

0.03\sqrt{n} = 1.645\sqrt{0.1(0.9)}

\sqrt{n} = \frac{1.645\sqrt{0.1(0.9)}}{0.03}

(\sqrt{n}})^2 = \left(\frac{1.645\sqrt{0.1(0.9)}}{0.03}\right)^2

n = 271 (rounded up).

More can be learned about the z-distribution at brainly.com/question/25890103

#SPJ1

3 0
1 year ago
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