Marcelo runs 3/4 of a mile in 1/8 of an hour. What is his unit rate in miles per hour?

Suppose that shoe sizes of American women have a bell-shaped distribution with a mean of 8.47 and a standard deviation of 1.5. U

alex41 [277]

Answer:

2.5% of American women have shoe sizes that are greater than 11.47.

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 8.47

Standard deviation = 1.5

The normal distribution is symmetric, which means that half the measures are below the mean, and half are above the mean.

What percentage of American women have shoe sizes that are greater than 11.47?

11.47 = 8.47 + 2*1.5

This means that 11.47 is two standard deviations above the mean.

Of those measures below the mean, none are greater than 11.47.

Of those measures above the mean, 95% are between the mean and 11.47, and 5% are above. So

0.5*0.05 = 0.025

2.5% of American women have shoe sizes that are greater than 11.47.

8 0

3 years ago

Initially a tank contains 10 liters of pure water. Brine of unknown (but constant) concentration of salt is flowing in at 1 lite

zhenek [66]

Answer:

Therefore the concentration of salt in the incoming brine is 1.73 g/L.

Step-by-step explanation:

Here the amount of incoming and outgoing of water are equal. Then the amount of water in the tank remain same = 10 liters.

Let the concentration of salt be a gram/L

Let the amount salt in the tank at any time t be Q(t).

$\frac{dQ}{dt} =\textrm {incoming rate - outgoing rate}$

Incoming rate = (a g/L)×(1 L/min)

=a g/min

The concentration of salt in the tank at any time t is = $\frac{Q(t)}{10}$ g/L

Outgoing rate = $(\frac{Q(t)}{10} g/L)(1 L/ min)$ $\frac{Q(t)}{10} g/min$

$\frac{dQ}{dt} = a- \frac{Q(t)}{10}$

$\Rightarrow \frac{dQ}{10a-Q(t)} =\frac{1}{10} dt$

Integrating both sides

$\Rightarrow \int \frac{dQ}{10a-Q(t)} =\int\frac{1}{10} dt$

$\Rightarrow -log|10a-Q(t)|=\frac{1}{10} t +c$ [ where c arbitrary constant]

Initial condition when t= 20 , Q(t)= 15 gram

$\Rightarrow -log|10a-15|=\frac{1}{10}\times 20 +c$

$\Rightarrow -log|10a-15|-2=c$

Therefore ,

$-log|10a-Q(t)|=\frac{1}{10} t -log|10a-15|-2$ .......(1)

In the starting time t=0 and Q(t)=0

Putting t=0 and Q(t)=0 in equation (1) we get

$- log|10a|= -log|10a-15| -2$

$\Rightarrow- log|10a|+log|10a-15|= -2$

$\Rightarrow log|\frac{10a-15}{10a}|= -2$

$\Rightarrow |\frac{10a-15}{10a}|=e ^{-2}$

$\Rightarrow 1-\frac{15}{10a} =e^{-2}$

$\Rightarrow \frac{15}{10a} =1-e^{-2}$

$\Rightarrow \frac{3}{2a} =1-e^{-2}$

$\Rightarrow2a= \frac{3}{1-e^{-2}}$

$\Rightarrow a = 1.73$

Therefore the concentration of salt in the incoming brine is 1.73 g/L

8 0

3 years ago

Round 66.3899872389 to the nearest thousandth

ddd [48]

Answer:

66.390

Step-by-step explanation:

6 0

3 years ago

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Select from the drop-down menus to correctly complete each statement. −20>−21 −19>−20 −21 is located to the of −20 on a nu

LekaFEV [45]

Answer:

left, right

Step-by-step explanation:

If you use the < or ≤ symbol to compare the numbers, their left-right order is the same as on the number line. Here, we would write the ordering as ...

-21 < -20 < -19

So, we see that ...

-21 is <em>left</em> of -20 on the number line

-19 is <em>right</em> of -20 on the number line

4 0

3 years ago

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Solve log (7x + 7) = 1. Round to the nearest thousandth if necessary. 0.429 2.333 1.143 0.875

frez [133]

Step-by-step explanation:

log(7·x + 7) = 1

7·x + 7 = 10^1

7·x = 10 - 7

x = 3/7 = 0.4285714285

3 0

3 years ago

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