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labwork [276]
3 years ago
14

Abdul runs 6 miles in 50 minutes. At the same rate, how many miles would he run in 35 minutes?

Mathematics
2 answers:
stepladder [879]3 years ago
5 0
\frac{6miles}{50min} = \frac{x miles}{35 min}  \\ 6*35 = 50*x \\ 210 =50x \\ 210/50=50x/50 \\ x=4.2
Abdul can run 4.2 miles in 35 minutes
lys-0071 [83]3 years ago
5 0
6 miles                    
--------                 ---------
50 minutes       35 minutes

35*6=210

210/50=4.2

FINAL ANSWER:
4.2




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the name Joe is very common at a school in one out of every ten students go by the name. If there are 15 students in one class,
kumpel [21]

Using the binomial distribution, it is found that there is a 0.7941 = 79.41% probability that at least one of them is named Joe.

For each student, there are only two possible outcomes, either they are named Joe, or they are not. The probability of a student being named Joe is independent of any other student, hence, the <em>binomial distribution</em> is used to solve this question.

<h3>Binomial probability distribution </h3>

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

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  • n is the number of trials.
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In this problem:

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  • There are 15 students in the class, hence n = 15.

The probability that at least one of them is named Joe is:

P(X \geq 1) = 1 - P(X = 0)

In which:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{15,0}.(0.1)^{0}.(0.9)^{15} = 0.2059

Then:

P(X \geq 1) = 1 - P(X = 0) = 1 - 0.2059 = 0.7941

0.7941 = 79.41% probability that at least one of them is named Joe.

To learn more about the binomial distribution, you can take a look at brainly.com/question/24863377

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XY bisects DF at point E. If DE=2y and EF=8y-3, find the value of DF​
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Answer:

Step-by-step explanation:

If XY bisects DF at point E, then DE = EF.

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DF = 5-3

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