All categories

3 years ago

max ran 40 miles in one week her friend maci ran 5\8 the distance that max ran. how many miles did maci run

Mathematics

1 answer:

Lostsunrise [7]3 years ago

5 0

EZ

all you have to do is divide 40 ÷ 5/8
and you get 1

You might be interested in

Which of the following is positive? -(-31), -(31), -31, 0-31

BartSMP [9]

It would be -(-31) because a negative and negative is a positive

3 0

3 years ago

Read 2 more answers

Square root of (93636)

Solnce55 [7]

Answer:

306

Step-by-step explanation:

mark brainliest :)

6 0

3 years ago

Evaluate 6 x (15 - 7) - 4 <br> A: 10<br> B: 24<br> C: 44<br> D: 79

Svet_ta [14]

First Of All Please Follow This
PEMDAS

P: ()
E:^
M:x
D:/
A:+
S:-

6 x (15 - 7) - 4
6 x (8) - 4
48 - 4
44

Answer: C

6 0

3 years ago

Read 2 more answers

Harper had $35 in her purse and found x dollars more on the couch. Cole had $41 in his wallet and spent twice as much money as H

yanalaym [24]

Answer (x is the money Harper found):

Equation: (35 + x) = (41 - 2x)
Solution: x = 2

I hope this helps!

8 0

2 years 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