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dangina [55]
3 years ago
14

Can U solve it plz explain fully !​

Mathematics
1 answer:
Gre4nikov [31]3 years ago
5 0

Answer:

(9 ^ 3) x 0.04 (4/99)

Step-by-step explanation:

This seems overwhelming, but after simplifying, it will look much easier

Numerator (First Row)

First, we will combine like terms and simplify the top row

(9 ^ <em>x</em> + 1) x 8 - (9 ^ <em>x</em> - 1) x 81

(9 ^ <em>x</em> + 1) - (9 ^ <em>x</em> - 1) x (8 x 81)

(9 ^ 2x) x 648

Denominator (Second Row)

Next, we can combine like terms and simplify the second row

(9 ^ <em>x</em>) x 66 - (9 ^ <em>x</em> - 2) x 243

(9 ^ <em>x</em>) - (9 ^ <em>x</em> - 2) x (66 x 243)

(9 ^ 2x + 2) x 16,038

Simplify

Our expression is now

(9 ^ 2x) x 648

(9 ^ 2x + 2) x 16,038

We can simplify it as

(9 ^ 3) x 0.04 (4/99)

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The perimeter of a rectangular field is 340 yards. If the length of the field is 91 yards, what is its width?
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79

Step-by-step explanation:

91 x 2 is 182. 340-182 is 158 divide by 2 and you get 79

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Find the value of (2^8 x 5^-5 x 19^0)^-2 x (5^-2 over 2^3)^4 x 2^28 simplify and show your work
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The yearly profits of a new small business, in hundreds of dollars, during its first 10 years can be modeled by p(x) = x3 + x2 –
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A machine that produces a major part for an airplane engine is monitored closely. In the past, 10% of the parts produced would b
nata0808 [166]

Answer:

With a .95 probability, the sample size that needs to be taken if the desired margin of error is .04 or less is of at least 216.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

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

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

The margin of error:

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

For this problem, we have that:

p = 0.1

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

With a .95 probability, the sample size that needs to be taken if the desired margin of error is .04 or less is

We need a sample size of at least n, in which n is found M = 0.04.

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

0.04 = 1.96\sqrt{\frac{0.1*0.9}{n}}

0.04\sqrt{n} = 0.588

\sqrt{n} = \frac{0.588}{0.04}

\sqrt{n} = 14.7

(\sqrt{n})^{2} = (14.7)^{2}

n = 216

With a .95 probability, the sample size that needs to be taken if the desired margin of error is .04 or less is of at least 216.

7 0
4 years ago
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