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DanielleElmas [232]
3 years ago
14

What is >>>>> -x<15-2x

Mathematics
1 answer:
Gemiola [76]3 years ago
7 0
-x
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Answer:

v olume=  \frac{1}{3}  \times  \: height \times surface \\ 1594.5 =  \frac{1}{3}  \times  {x}^{2}  \times 15.8 \\  {x}^{2} = (1594.5 \times 3) \div 15.8 = 302.753 \\ x = 17.399

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Loan for Vehicle Loan APR Payment #1 $20,000 for 66 months 2.17% $322 #2 $25,000 for 78 months 3.38% $357 #3 $30,000 for 84 mont
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The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e
irakobra [83]

Answer:

We need a sample size of at least 383.

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 is:

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

85% confidence level

So \alpha = 0.15, z is the value of Z that has a pvalue of 1 - \frac{0.15}{2} = 0.925, so Z = 1.44.

How large a sample would be required in order to estimate the fraction of tenth graders reading at or below the eighth grade level at the 85% confidence level with an error of at most 0.03

We need a sample size of at least n.

n is found with M = 0.03, \pi = 0.21

Then

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

0.03 = 1.44\sqrt{\frac{0.21*0.79}{n}}

0.03\sqrt{n} = 1.44\sqrt{0.21*0.79}

\sqrt{n} = \frac{1.44\sqrt{0.21*0.79}}{0.03}

(\sqrt{n})^{2} = (\frac{1.44\sqrt{0.21*0.79}}{0.03})^{2}

n = 382.23

Rounding up

We need a sample size of at least 383.

6 0
3 years ago
Total freshman = 58. Number of freshman who prefer cheese toppings= 14. Determined the probability that a freshman prefer cheese
Yakvenalex [24]

Answer:

The probability that a freshman prefer cheese toppings  is   0.241

Step-by-step explanation:

Probability of any event E =  \frac{\textrm{Number of favorable outcomes}}{\textrm{Total number of outcomes}}

Here, let E   :  Event of choosing cheese toppings

So, the number of favorable outcomes = 14

Total number of outcomes = 58

So, P(E) =  \frac{\textrm{Number of students who like cheese toppings}}{\textrm{Total number of students}}

or, P(E) = \frac{14}{58} = 0.241379

So,the probability that a freshman prefer cheese toppings  is   0.241379

Rounding of 0.241379  to the <u>nearest thousandth</u>, we get

Here , in 0.241379 thousandth digit is 3, and 3 < 5,

so the value of P(E) = 0.241

4 0
3 years ago
Another question (picture)
ASHA 777 [7]


\left( 16 x^8 y^{-12} \right)^{\frac 1 2} = 16^{\frac 1 2} (x^8)^{\frac 1 2} (y^{-12})^{\frac 1 2} = 4 x^4 y^{-6} = \dfrac{4 x^4}{y^6}


Third choice



8 0
3 years ago
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