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3 years ago

What would this be simplified (x^6)^5*(x^4)^7

Mathematics

1 answer:

SVEN [57.7K]3 years ago

8 0

Simplified form would be x^58

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Please awnser this is due in the morning

Anna007 [38]

where is the question

5 0

3 years ago

After calculating the sample size needed to estimate a population proportion to within 0.05, you have been told that the maximum

Svetlanka [38]

Answer: 40000

Step-by-step explanation:

The formula to find the sample size is given by :-

$n=p(1-p)(\dfrac{z_{\alpha/2}}{E})^2$ , where p is the prior estimate of the population proportion.

Here we can see that the sample size is inversely proportion withe square of margin of error.

i.e. $n\ \alpha\ \dfrac{1}{E^2}$

By the equation inverse variation, we have

$n_1E_1^2=n_2E_2^2$

Given : $E_1=0.05$ $n_1=1000$

$E_2=0.025$

Then, we have

$(1000)(0.05)^2=n_2(0.025)^2\\\\\Rightarrow\ 2.5=0.000625n_2\\\\\Rightarrow\ n_2=\dfrac{2.5}{0.000625}=4000$

Hence, the sample size will now have to be 4000.

6 0

3 years ago

74 divided by 2 but long division

Olin [163]

Answer:

Step-by-step explanation:

74/2

8 0

3 years ago

An article is bought for Rs. 125 what is the profit percentage?

Aneli [31]

Step-by-step explanation:

Solution given;

cost price=Rs125

profit%=?

we have

profit%=[Selling price-cost price]/cost price×100%

=[selling price-Rs.125]/Rs 125×100% is your answer

7 0

3 years ago

Set of data with 350,000 numbers is normally

Rudiy27

Step-by-step explanation:

given a normal distribution with the given parameters the probability (= the % of the area of the distribution curve) for a number to be between 203 and 1803 is

0.9987

so, 99.87% of all numbers are expected to be in that range.

for 350,000 numbers that means

350,000×0.9987 = 349,545 numbers are expected to be between 203 and 1803.

7 0

2 years ago