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3 years ago

Please help Solve for r. In fraction form please

Mathematics

1 answer:

klio [65]3 years ago

5 0

Answer:

28571/1000

Step-by-step explanation:

r/5 4/7

Cross Multiply to get

r7 and 20

divide both sides by 7

r = 2.8571

Since that is in the thousands place the answer is 28571/1000

You might be interested in

Suppose that the times required for a cable company to fix cable problems in its customers’ homes are uniformly distributed betw

vampirchik [111]

Answer:

a) 60% probability that a randomly selected cable repair visit will take at least 50 minutes

b) 60% probability that a randomly selected cable repair visit will take at most 55 minutes.

c) The expected length of the repair visit is 52.5 minutes.

d) The standard deviation is 7.22 minutes.

Step-by-step explanation:

An uniform probability is a case of probability in which each outcome is equally as likely.

For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.

The probability that we find a value X lower than x is given by the following formula.

$P(X \leq x) = \frac{x - a}{b-a}$

For this problem, we have that:

Uniformly distributed between 40 and 65 minutes, so $a = 40, b = 65$

(a) What is the probability that a randomly selected cable repair visit will take at least 50 minutes?

Either it takes less than 50 minutes, or it takes at least 50 minutes. The sum of the probabilities of these events is decimal 1. So

$P(X < 50) + P(X \geq 50) = 1$

The intervals can be open or closed, so

$P(X < 50) = P(X \leq 50)$

We have that

$P(X \geq 50) = 1 - P(X < 50)$

$P(X < x) = \frac{50 - 40}{65 - 40} = 0.4$

$P(X \geq 50) = 1 - P(X < 50) = 1 - 0.4 = 0.6$

60% probability that a randomly selected cable repair visit will take at least 50 minutes

(b) What is the probability that a randomly selected cable repair visit will take at most 55 minutes?

This is $P(X \leq 55)$

$P(X \leq 55) = \frac{55 - 40}{65 - 40} = 0.6$

60% probability that a randomly selected cable repair visit will take at most 55 minutes.

(c) What is the expected length of the repair visit?

The mean of the uniform distribution is:

$M = \frac{a+b}{2}$

$M = \frac{40+65}{2} = 52.5$

The expected length of the repair visit is 52.5 minutes.

(d) What is the standard deviation?

The standard deviation of the uniform distribution is

$S = \sqrt{\frac{(b-a)^{2}}{12}}$

$S = \sqrt{\frac{(65-40)^{2}}{12}}$

$S = 7.22$

The standard deviation is 7.22 minutes.

5 0

3 years ago

I WILL GIVE BRAINLIEST AND 10 POINTS IF ANSWERED QUICKLY!!!!!!!!!!!!!

vladimir1956 [14]

Answer: A=100(12)t5.27

Step-by-step explanation:

Assume / and ^ got lost from eqns.

A(t) = initial × (1/2)^(t/halflife)

So initial = 100, halflife = 5.27,

A=100(12)t5.27 means A=100(1/2)^(t/5.27)

so that A is 50 when t is 5.27

7 0

3 years ago

X + (x − 14) = 40. little confused

ikadub [295]

Answer:

27is answer

Step-by-step explanation:

x+ x - 14 = 40

or, 2x = 40+ 14

or, 2x = 54

or, x = <u>5</u><u>4</u>

or, x= 27

7 0

3 years ago

Will give Brainliest!!!

professor190 [17]

Answer:

5 x T = L

Step-by-step explanation:

8 0

3 years ago

IM DOIGN A MATH TEST AND I NEED HELP RN

Vladimir79 [104]

Answer:

its 43

Step-by-step explanation:

i had a test with the same question, i promise its right :)

6 0

4 years ago