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lilavasa [31]
3 years ago
11

Kinda stuck please help would appreciate it :)

Mathematics
2 answers:
V125BC [204]3 years ago
8 0

<em>Answer,</em>

<em>2w - 5 = 8</em>


<u><em>Hope this helps :-)</em></u>

Alla [95]3 years ago
3 0
2(n-5)=8
Difference is subracting
n=number
subtract 5 from n for the difference
now multiply it by two
when a problem says "twice the difference of..." it means multiply by two. Put that in parenthesis to keep it separate, then(if you want to go a step further) distribute.
2n-10=8
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The total scores on the Medical College Admission Test (MCAT) in 2013 follow a Normal distribution with mean 25.3 and standard d
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In a normal distribution, the median is the same as the mean (25.3). The first quartile is the value of Q_1 such that

\mathbb P(X

You have

\mathbb P(X

For the standard normal distribution, the first quartile is about z\approx-0.6745, and by symmetry the third quartile would be z\approx0.6745. In terms of the MCAT score distribution, these values are

\dfrac{Q_1-25.3}{6.5}=-0.6745\implies Q_1\approx20.9
\dfrac{Q_3-25.3}{6.5}=0.6745\implies Q_3\approx29.7

The interquartile range (IQR) is just the difference between the two quartiles, so the IQR is about 8.8.

The central 80% of the scores have z-scores \pm z such that

\mathbb P(-z

That leaves 10% on either side of this range, which means

\underbrace{\mathbb P(-z

You have

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Converting to MCAT scores,

-1.2816=\dfrac{x_{\text{low}}-25.3}{6.5}\implies x_{\text{low}}\approx17.0
1.2816=\dfrac{x_{\text{high}}-25.3}{6.5}\implies x_{\text{high}}\approx33.6

So the interval that contains the central 80% is (17.0,33.6) (give or take).
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3 years ago
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Step-by-step explanation:

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The value of a car bought new for $28900 decreases 15% each year. Identify the function for the value of the car. Does the funct
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Answer:

<u>V (t) = 28,900 - 4,335t</u>

<u>The function clearly represent a decay</u>

Step-by-step explanation:

1. Let's review the information given to us to answer the question correctly:

Value of the car = $ 28,900

Annual depreciation = 15% = 0.15

2. Identify the function for the value of the car. Does the function represent growth or decay?

Let y represent the value of the car after t years of utilization, let p the price of the car and d, the annual depreciation, therefore:

V (t) = p - (d * t * p)

Replacing with the values we know:

V (t) = 28,900 - (0.15 * t * 28.900)

V (t) = 28,900 - (0.15 * t * 28.900)

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