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2 years ago

jose has a board that is 15 ft long he wants to cut it into 3/4 ft long how many pieces will he have ?

Mathematics

1 answer:

IrinaK [193]2 years ago

7 0

Answer: 20

Step-by-step explanation: the equation would be 15 / 3/4, or 15/1 / 3/4 (the / meaning divide). You would divide it by multiplying the reciprocal of 3/4 (4/3) and 15(/1) , which you would get 20.

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The graph below shows the speed of a car that is driven through a town and then on a major highway. During which of the followin

Varvara68 [4.7K]

Considering the graph of the velocity of the car, it is found that the interval in which it was stopped at a traffic light was:

Between 3 and 4 minutes.

<h3>When is a car stopped at a traffic light?</h3>

When a car is stopped at a traffic light, the car is not moving, that is, it's velocity is of zero.

In this problem, the graph gives the <u>velocity as a function of time</u>, and it is at zero between 3 and 4 minutes, hence the interval in which it was stopped at a traffic light was:

Between 3 and 4 minutes.

More can be learned about the interpretation of the graph of a function at brainly.com/question/3939432

#SPJ1

4 0

1 year ago

The total claim amount for a health insurance policy follows a distribution with density function 1 ( /1000) ( ) 1000 x fx e− =

gizmo_the_mogwai [7]

Answer:

the approximate probability that the insurance company will have claims exceeding the premiums collected is $\mathbf{P(X>1100n) = 0.158655}$

Step-by-step explanation:

The probability of the density function of the total claim amount for the health insurance policy is given as :

$f_x(x) = \dfrac{1}{1000}e^{\frac{-x}{1000}}, \ x> 0$

Thus, the expected total claim amount $\mu$ = 1000

The variance of the total claim amount $\sigma ^2 = 1000^2$

However; the premium for the policy is set at the expected total claim amount plus 100. i.e (1000+100) = 1100

To determine the approximate probability that the insurance company will have claims exceeding the premiums collected if 100 policies are sold; we have :

P(X > 1100 n )

where n = numbers of premium sold

$P (X> 1100n) = P (\dfrac{X - n \mu}{\sqrt{n \sigma ^2 }}> \dfrac{1100n - n \mu }{\sqrt{n \sigma^2}})$

$P(X>1100n) = P(Z> \dfrac{\sqrt{n}(1100-1000}{1000})$

$P(X>1100n) = P(Z> \dfrac{10*100}{1000})$

$P(X>1100n) = P(Z> 1) \\ \\ P(X>1100n) = 1-P ( Z \leq 1) \\ \\ P(X>1100n) =1- 0.841345$

$\mathbf{P(X>1100n) = 0.158655}$

Therefore: the approximate probability that the insurance company will have claims exceeding the premiums collected is $\mathbf{P(X>1100n) = 0.158655}$

4 0

2 years ago

Is 17 greater then 0.17

Ratling [72]

Yes, since 17 is larger than 1 and 0.17 is less than 1.

4 0

3 years ago

How many numbers between 325 and 564 end with 0?

IgorLugansk [536]

24 numbers that end with 0 between 325 and 564

6 0

3 years ago

There is an "X" drawn on a sheet of paper. One of the smaller angles created is 57 degrees. What are the other 3 angle measures?

sukhopar [10]

In total the answer has to be 180 degrees so the other 2 angels would be any 2 numbers that when you add them and 57 together you get 180

7 0

2 years ago