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2 years ago

Yo someone hellp me and if you see this please click!

Mathematics

2 answers:

Vinvika [58]2 years ago

4 0

Answer:

3/4

Step-by-step explanation:

weqwewe [10]2 years ago

3 0

Answer: 3/4

Step-by-step explanation:

In decimal form, 2/5 is .4

In decimal form, 7/8 is .875.

In decimal form, 3/4 is .75.

In decimal form, 15/16 is .9375.

In decimal form, 1/3 is .3333.

In decimal form, 3/8 is .375.

3/4 is the only one in between.

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What are all the names that apply to <br> √24

Kisachek [45]

Answer: Square Root of 24

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2 years ago

Cual es la relacion entre <a y <b?

RideAnS [48]

Answer:

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2 years ago

Helen walk 2 3/4 miles per day. Bill run's 10 miles every 4 days. In 4 days, who runs the longest distance

Gennadij [26K]

Helen runs the longest distance in 4 days.

Helen (In 4 days): 2.75 x 4 = 11

Bill (In 4 days): 10

11 (Helen) > 10 (Bill), so therefore Helen runs the longest distance in 4 days.

5 0

3 years ago

June is driving from Brookline, Massachusetts to Brooklyn, new york. The cities are 200 miles apart. Pretend June lives in a wor

I am Lyosha [343]

It took June 4 hours to get from Brookline to Brooklyn.

<h3><u>Time calculation</u></h3>

Given that June is driving from Brookline, Massachusetts to Brooklyn, New York, and the cities are 200 miles apart, and she can drive at a constant speed of 50mph the entire way, to determine the function that solves the time June takes to complete the route, the following calculation must be made:

C / Constant speed = T
200 / 50 = T
4 = T

So it took June 4 hours to get from Brookline to Brooklyn.

Learn more about time calculation in brainly.com/question/26707195

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