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svlad2 [7]
3 years ago
11

Rewrite 1 / 6 barrels over 2 over 5 hours as a unit rate

Mathematics
1 answer:
Akimi4 [234]3 years ago
5 0

Answer:

5/12 barrel/hour

Step-by-step explanation:

 Type this as a fraction:

  (1/6) barrel

--------------------

  (2/5) hours

Division by 2/5 is equivalent to multiplication by 5/2:

1                5

--- barrel * -------------- = 5/12 barrel/hour

6                2  hours

Alternatively, this problem could be done using decimal fractions:

0.1667 barrel

-------------------- = 0.416666 barrel/hour  =  5/12 barrel/hour

0.4 hours

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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
Elden [556K]
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

\mathbb P(Z

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).
7 0
2 years ago
To solve the system of equations below, chen isolated x2 in the first equation and then substituted it into the second equation.
Ann [662]
Im pretty sure it should be 

25-y^2/16 -y^2/9=1
8 0
3 years ago
Read 2 more answers
A supervisor records the repair cost for 22 randomly selected VCRs. A sample mean of $75.50 and standard deviation of $18.07 are
VashaNatasha [74]

Answer:

The t value for 99% CI for 21 df is 2.831.

The critical value that should be used in constructing the confidence interval is (64.593, 86.407).

Step-by-step explanation:

Now the sample size is less than 30 and also population standard deviation is not known.

Then we will use t distribution to find CI

t value for 99% CI for 21 df is TINV(0.01,21)=2.831

The margin of error is E=t\times\frac{s}{\sqrt{n}}\\\\=2.831\times\frac{18.07}{\sqrt{22}}\\\\=10.907

Hence CI isCI=\overline{x} \pm E\\\\ =75.50 \pm 10.907\\\\=(64.593,86.407 )

8 0
2 years ago
Annabelle owns a small business selling clothing. She knows that in the last week 56 customers paid cash, 26 customers used a de
frosja888 [35]

Answer

23/51

Step-by-step explanation:

The probability of the next customer paying with a debit or credit card is =23/51

5 0
1 year ago
C = 1 /2Rt solve for R
Vladimir [108]

C = 1/2 RT

1. C/2 = RT

2. C/2 * 1 / T = R

or

R = C / 2 * 1 / T

7 0
3 years ago
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