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3 years ago

Nita runs 6 miles in 54 minutes. what is the rate in minutes per mile

Mathematics

2 answers:

drek231 [11]3 years ago

8 0

Answer:

a 6x54=324 mile per minute

Step-by-step explanation:

bulgar [2K]3 years ago

7 0

Answer:

The answer is 324 mile per minute.

Step-by-step explanation:

54 x 6 = 324 minutes/mile. I hope you got your answer :) Nice question, I just figured out this question :)

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How do you write 3.25 in simplest form

dedylja [7]

3.25 would be 3 1/4 because,

Step 1: You would rewrite the decimal number as a fraction with 1 in the denominator. So, 3.25 = 3.24/1

Step 2: Multiply it by 1 to eliminate 2 decimals places, so multiply the top and bottom by 10^2 = 100
3.25/1 x 100/100 = 325/100
Step 3: Now you find the greatest common factor, also known as GCF, of 325 and 100 if it exits then you reduce the fraction by dividing both numerator and denominator by it. GCF = 25
3.25/100 divided by 25/25 = 13/4
Now you would simplify it to make a proper fraction getting your answer
3 1/4
Does that make sense?

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3 years ago

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A scientist was in a submarine below sea level, studying ocean life. Over the next ten minutes, she ascended 63.9 feet. How many

kati45 [8]

Answer:

-85.9 feet below sea level.

Step-by-step explanation:

3 0

2 years ago

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PLEASE HELP 12 POINTS

Assoli18 [71]

Answer:

5:5 (first box, pencils to pens)

7:3 (second box, coloured pencils to crayons)

The probability of picking a pen (1st box): 5/10

The probability of picking a crayon (2nd box): 3/10

Probability of picking both: 5/10*3/10 = 15/100

3 0

3 years ago

A foreign student club lists as its members 2 Canadians, 3 Japanese, 5 Italians, and 2 Germans. If a committee of 4 is selected

Fittoniya [83]

Answer:

(a) The probability that the members of the committee are chosen from all nationalities $=\frac{4}{33}$ =0.1212.

(b)The probability that all nationalities except Italian are represent is 0.04848.

Step-by-step explanation:

Hypergeometric Distribution:

Let $x_1$ , $x_2$ , $x_3$ and $x_4$ be four given positive integers and let $x_1+x_2+x_3+x_4= N$ .

A random variable X is said to have hypergeometric distribution with parameter $x_1$ , $x_2$ , $x_3$ , $x_4$ and n.

The probability mass function

$f(x_1,x_2.x_3,x_4;a_1,a_2,a_3,a_4;N,n)=\frac{\left(\begin{array}{c}x_1\\a_1\end{array}\right)\left(\begin{array}{c}x_2\\a_2\end{array}\right) \left(\begin{array}{c}x_3\\a_3\end{array}\right) \left(\begin{array}{c}x_4\\a_4\end{array}\right) }{\left(\begin{array}{c}N\\n\end{array}\right) }$

Here $a_1+a_2+a_3+a_4=n$

${\left(\begin{array}{c}x_1\\a_1\end{array}\right)=^{x_1}C_{a_1}= \frac{x_1!}{a_1!(x_1-a_1)!}$

Given that, a foreign club is made of 2 Canadian members, 3 Japanese members, 5 Italian members and 2 Germans members.

$x_1$ =2, $x_2$ =3, $x_3$ =5 and $x_4$ =2.

A committee is made of 4 member.

N=4

(a)

We need to find out the probability that the members of the committee are chosen from all nationalities.

$a_1$ =1, $a_2$ =1, $a_3$ =1 , $a_4$ =1, n=4

The required probability is

$=\frac{\left(\begin{array}{c}2\\1\end{array}\right)\left(\begin{array}{c}3\\1\end{array}\right) \left(\begin{array}{c}5\\1\end{array}\right) \left(\begin{array}{c}2\\1\end{array}\right) }{\left(\begin{array}{c}12\\4\end{array}\right) }$

$=\frac{2\times 3\times 5\times 2}{495}$

$=\frac{4}{33}$

=0.1212

(b)

Now we find out the probability that all nationalities except Italian.

So, we need to find out,

$P(a_1=2,a_2=1,a_3=0,a_4=1)+P(a_1=1,a_2=2,a_3=0,a_4=1)+P(a_1=1,a_2=1,a_3=0,a_4=2)$

$=\frac{\left(\begin{array}{c}2\\2\end{array}\right)\left(\begin{array}{c}3\\1\end{array}\right) \left(\begin{array}{c}5\\0\end{array}\right) \left(\begin{array}{c}2\\1\end{array}\right) }{\left(\begin{array}{c}12\\4\end{array}\right) }+\frac{\left(\begin{array}{c}2\\1\end{array}\right)\left(\begin{array}{c}3\\2\end{array}\right) \left(\begin{array}{c}5\\0\end{array}\right) \left(\begin{array}{c}2\\1\end{array}\right) }{\left(\begin{array}{c}12\\4\end{array}\right) }$ $+\frac{\left(\begin{array}{c}2\\1\end{array}\right)\left(\begin{array}{c}3\\1\end{array}\right) \left(\begin{array}{c}5\\0\end{array}\right) \left(\begin{array}{c}2\\2\end{array}\right) }{\left(\begin{array}{c}12\\4\end{array}\right) }$