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3 years ago

A relay team of 4 people runs 3 miles. If each person runs the same distance, how many mile does each person run?

Mathematics

2 answers:

Talja [164]3 years ago

7 0

Given

A relay team of 4 people runs 3 miles .

If each person runs the same distance.

Find out how many mile does each person run .

To prove

As given

4 people runs 3 miles .

If each person runs the same distance.

Than

$Each\ person\ runs\ the\ distance = \frac{3}{4}\ miles$

$Therefore\ each\ person\ runs\ the\ distance\ is\ \frac{3}{4}\ miles.$

$Therefore\ each\ person\ runs\ the\ distance\ is\ 0.75\ miles.$

Lera25 [3.4K]3 years ago

3 0

Answer:

Miles ran by each person = 0.75

Explanation:

Total distance traveled = 3 miles.

Total number of persons in relay = 4

Distance traveled by each person = Total distance/ Total number of persons

= 3 / 4 = 0.75 miles per person.

Miles ran by each person = 0.75

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3 years ago

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Answer:

2:55 P.M.

Step-by-step explanation:

First, add the amount of time Jim played outside; 45+55=100

100 minutes is equal to 1 hour 40 minutes.

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Subtract 1 hour 40 minutes from 4:35 P.M. to then get 2:55 P.M.

To confirm your answer, you can add back the 1 hour and 40 minutes to 2:55 P.M. and you'll get 4:35 P.M.; back with what you started with.

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3 years ago

A manufacturer knows that on average 20% of the electric toasters produced require repairs within 1 year after they are sold. Wh

Effectus [21]

Answer:

(a) The value of x is 5.

(b) The value of y is 15.

Step-by-step explanation:

Let the random variable X represent the number of electric toasters produced that require repairs within 1 year.

And the let the random variable Y represent the number of electric toasters produced that does not require repairs within 1 year.

The probability of the random variables are:

P (X) = 0.20

P (Y) = 1 - P (X) = 1 - 0.20 = 0.80

The event that a randomly selected electric toaster requires repair is independent of the other electric toasters.

A random sample of n = 20 toasters are selected.

The random variable X and Y thus, follows binomial distribution.

The probability mass function of X and Y are:

$P(X=x)={20\choose x}(0.20)^{x}(1-0.20)^{20-x}$

$P(Y=y)={20\choose y}(0.20)^{20-y}(1-0.20)^{y}$

(a)

Compute the value of x such that P (X ≥ x) < 0.50:

$P (X \geq x) < 0.50\\\\1-P(X\leq x-1)$

Use the Binomial table for n = 20 and p = 0.20.

$0.411=\sum\limits^{3}_{x=0}[b(x,20,0.20)]$

The least value of x that satisfies the inequality P (X ≥ x) < 0.50 is:

x - 1 = 4

x = 5

Thus, the value of x is 5.

(b)

Compute the value of y such that P (Y ≥ y) > 0.80:

$P (Y \geq y) >0.80\\\\P(Y\leq 20-y)>0.80\\\\P(Y\leq 20-y)>0.80\\\\\sum\limits^{20-y}_{y=0}[{20\choose y}(0.20)^{20-y}(1-0.20)^{y}]>0.80$

Use the Binomial table for n = 20 and p = 0.20.

$0.630=\sum\limits^{4}_{y=0}[b(y,20,0.20)]$

The least value of y that satisfies the inequality P (Y ≥ y) > 0.80 is:

20 - y = 5

y = 15

Thus, the value of y is 15.

3 0

3 years ago