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4 years ago

URGENTS PLLLLLSSSSSSSSS

Mathematics

2 answers:

kobusy [5.1K]4 years ago

8 0

A. 12
You would need to divide 300/25 because 25% of people like vanilla. The answer is 12 people.

Arlecino [84]4 years ago

5 0

Just divide 300 by 4 since 25% is 1/4
300/4 = 75
D. 75
Hope this helps!

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Complete each statement to describe the relationship between gallons and liters.

maw [93]

Conversion
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To convert gallons to liters multiply gallons by 3.785

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To convert liters to gallons multiply liters by 0.264172

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2 years ago

The U.S census bureau reported that 9 out of every 20 joggers at female. On a trail there were 230 joggers on July 4. Approximat

hodyreva [135]

Answer:

Approximately 104 joggers are female

Step-by-step explanation:

The U.S census bureau reported that 9 out of every 20 joggers at female.

Let x be the number of females out of 230 joggers.

Rewrite it as

$\begin{array}{cc}\text{Female}&\text{All joggers}\\9&20\\x&230\end{array}$

Now we can write a proportion:

$\dfrac{9}{x}=\dfrac{20}{230}$

Cross multiply:

$20x=9\cdot 230\\ \\2x=9\cdot 23\ [\text{Divided by 10}]\\ \\x=\dfrac{9\cdot 23}{2}=\dfrac{207}{2}=103.5\approx 104$

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3 years ago

A man and a woman agree to meet at a cafe about noon. If the man arrives at a time uniformly distributed between 11:40 and 12:10

kompoz [17]

Answer:

0.3888

Step-by-step explanation:

To solve the exercise we need a neutral time, so we define at 12:00 as time 0.

If the man arrives in a uniformly distributed time between 11:40 and 12:10, the time is distributed from -20 to 10.

If the woman arrives in a time evenly distributed between 11:45 and 12:20, the time is distributed from -15 to 20.

Assuming that each one arrives at 12: X and 12: Y for men and women, then the space of our time is defined as

Man [-20,10]

Woman [-15,20]

So,

$f_x(x)=\frac{1}{10-(-20)}=\frac{1}{30}\\f_y(y)=\frac{1}{20-(-15)}=\frac{1}{35}$

Then,

$f_{xy}(x,y)=f_x(x)f_y(y)=\frac{1}{30} \frac{1}{35} = \frac{1}{1050}$

The probability of finding an arrival less than 5 minutes is

$P(|X-Y|\leq 5) = \int\limit^20_{-15} \int\limit^{x+5}_{x-5} f(x,y)dydx$

$P(|X-Y|\leq 5)= \frac{1}{1050}\int\limit^{20}_{-15} y|^{x+5}_{x-5}dx$

$P(|X-Y|\leq 5)= \frac{35*10}{900}=0.3888$

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Answer:

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Step-by-step explanation:

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3 years ago

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