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4 years ago

How long would it take for a person jog?

1 answer:

dsp734 years ago

8 0

This depends on a lot of factors. A persons weight, height, etc. It also depends on how many calories they want to burn or if they are just looking to run for fun. A lot of different factors play into a persons time to jog.

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5 time 5 to the power of 2 simplified

Maksim231197 [3]

Im not sure wym but i think the answer is 20

3 0

3 years ago

Read 2 more answers

( 10.2 -1)+(5.2 48 -8)

never [62]

Answer:35

Step-by-step explanation:

6 0

3 years ago

Find h.<br> 10 in.<br> h<br> h = [?] in.<br> 4 in.

aleksklad [387]

Answer:

h = sqrt(96)

Step-by-step explanation:

We can use the pythagorean theorem

a^2 + b^2 = c^2 where a and b are the legs and c is the hypotenuse

2^2 + h^2 = 10^2

4 + h^2 = 100

h^2 = 100-4

h^2 = 96

Taking the square root of each side

h = sqrt(96)

8 0

3 years ago

A computer retail store has 1414 personal computers in stock. A buyer wants to purchase 33 of them. Unknown to either the retail

Lorico [155]

Answer:

a) 364 ways

b) 45.33% probability that exactly one of the computers will be defective.

c) 54.67% probability that at least one of the computers selected is defective.

Step-by-step explanation:

The computers are chosen without replacement, so we use the hypergeometric distribution to solve this question.

Hypergeometric distribution:

The probability of x sucesses is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

In this question:

14 computers, so N = 14.

3 defective, so k = 3.

3 will be purchesed, so n = 3.

A) In how many different ways can the 3 computers be chosen?

3 from a set of 14. So

$C_{14,3} = \frac{14!}{3!(14-3)!} = 364$

364 ways

B) What is the probability that exactly one of thecomputers will be defective?

This is P(X = 1).

$P(X = 1) = h(1,14,3,3) = \frac{C_{3,1}*C_{11,2}}{C_{14,3}} = 0.4533$

45.33% probability that exactly one of the computers will be defective.

C) What is the probability that at least one of the computers selected is defective?

Either none is, or at least one is. The sum of the probabilities of these events is 1. So

$P(X = 0) + P(X \geq 1) = 1$

We want $P(X \geq 1)$ . Then

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

In which

$P(X = 0) = h(0,14,3,3) = \frac{C_{3,0}*C_{11,3}}{C_{14,3}} = 0.4533$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.4533 = 0.5467$

54.67% probability that at least one of the computers selected is defective.

5 0

3 years ago

You see a used car you wish to buy. The dealer quotes you a price of $1,595. You have a Blue Book quotation of $1,435 for the sa

Arada [10]

1.1 %. Take $1435*%=1595

so 1595/1435=1.1114982%

5 0

4 years ago

Read 2 more answers