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3 years ago

PLEASE HELP ⁉️ POR FAVOR SOMEONE IM DESPERATE tHIS IS DUE IN THE mORiNG!

Mathematics

1 answer:

Lerok [7]3 years ago

4 0

Answer:

Just in case you want me to answer all the questions, I will

Step-by-step explanation:

Question 1): answer is 25

Question 2): answer is 24

Question 3): answer is 100%

24/25 = 96/100

Box Question 1): 21/25 = 84/100

Box Question 2): 15/20 = 75/100

Box Question 3): 47/50 = 94/100

Box Question 4): 28/30 = 93.<u>3</u>/100

Box Question 5): 19/45 = 42.<u>2</u>/100

Box Question 6): 3/4 = 75/100

If a digit is underlined, put a line ONLY over the underlined digit, it is repeating

I'm not completely sure if this is correct, but I really think it is..

I hope this helps!

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How to approximate to 3 significant figures

kondaur [170]

Answer:

We round a number to three significant figures in the same way that we would round to three decimal places. We count from the first non-zero digit for three digits. We then round the last digit. We fill in any remaining places to the right of the decimal point with zeros.

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2 years ago

A number b increased by 3 is greater than or equal to -26

photoshop1234 [79]

A number b increased by 3 is greater than or equal to -26:

$b+3\geq-26\qquad\text{subtract 3 from both sides}\\\\\boxed{b\geq-29}$

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

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2 years ago

What two numbers multiplied equals 75 and add up to 10

HACTEHA [7]

So I don't know if you typed the question in wrong but if you solve simultaneously for:

xy= 75
x + y= 10

you will get the linked image I got from a graphics calculator program

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3 years ago

1. Choose the problem that matches the number line solution

Taya2010 [7]

Answer:

2×3/5

..................

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2 years ago

Read 2 more answers