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3 years ago

Please answer this friends I am stuck ?

Mathematics

1 answer:

goblinko [34]3 years ago

4 0

Answer:

<u>10°C average temperature</u>

Step-by-step explanation:

Average Temperature = Lowest + 17

= -7 + 17

= 17 - 7

= <u>10°C average temperature</u>

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3 years ago

Read 2 more answers

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

a small rock falling from the top of a 120 ft tall building with an intial downward velocity of -30 ft/sec is modeled by the equ

Effectus [21]

Answer:

The rock remains in the air for t = 3.832 s

Step-by-step explanation:

To easily solve this equation, we can graph the equation with a calculator and see the interval for which the rock remains in the air.

(There is a minor problem with the equation, the last term should be 120 to model the 120 ft building)

h(t)= -16t^2 -30t + 120

Please see attached picture for the graph.

The time it takes for the ball to hit ground can be found by making

h(t) = -16t^2 -30t + 120 = 0

In the graph this point is equal to

t = 3.832 s

h(t) = 0

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3 years ago