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3 years ago

Which of the following ordered pairs is the solution to the system of linear equations?

1 answer:

3 0

Solve for the first variable in one the equations then substitute the result into the other equation so the answer is (2,5)

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Samara was playing a game. She wrote down this number sentence to find the total number of points for 11 games in a row. 5 + 7 +

Firdavs [7]

Answer:55

Step-by-step explanation: Whenever you see the word "sum" it means to add. So, 5+7+9+3+1+3+9+8+5+2+3=55

8 0

4 years ago

Write the expression that represents the phrase, "one third the sum of a number and 10." Use x as the variable

agasfer [191]

Answer:

$\frac{x+10}{3}$ is the expression that represents the phrase.

Step-by-step explanation:

Given:

The phrase, "one third the sum of a number and 10."

Let the number be x

∴ The expression will be= $\frac{1}{3}\times (x+10)= \frac{x+10}{3}$

3 0

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

4 0

3 years ago

Last year, Nick went bowling several times and earned an average score

Norma-Jean [14]

Answer: 150%

Explanation:

1. Turn it into a fraction. (123/82)

2. Divide the numerator (123) by the denominator (82). (123÷82=1.5)

3. Multiply the decimal (1.5) by 100 and add “%” at the end to make it a percentage. (1.5x100=150)(150=150%)

4 0

3 years ago

A pyramid with a square base has a volume of 912 cubic feet and a height of 19 feet. Find the side length of the square base

Anit [1.1K]

The answer that you are looking for is 32

6 0

3 years ago