<span>There is a 20% (or 20/100chance) it will rain on Thursday and there is a 40% (or 40/100) chance it will rain on Friday. In order to find out the probability of it raining both days, we need to multiply these two values. (40/100)(20/100)= 800/10,000= 0.08. We move the decimal over two places to get a percent, so we get an 8% chance it will rain on both Thursday and Friday.</span>

8 0

4 years ago

Can someone please help me

Vsevolod [243]

The restrictions for the equation is that the denominator can not be zero. So the restrictions for x would be what values make the denominator zero.

7x^2 + 6x = 0
factor
x(7x + 6) = 0
multiply any number by 0 and you get 0. So either x = 0 or 7x+6 = 0
since there's a x in the numerator x/x = 1 so this will not be a restriction. Then the only restriction is:
7x+6 = 0
7x = -6
x = -6/7

5 0

3 years ago

8. Loren already knows that he will have $500,000 when he retires. If he sets up a payout

Ulleksa [173]

The annuity could provide $221.19 each month

Step-by-step explanation:

The ANNUITY FORMULA is $P_{t}=\frac{d[(1+\frac{r}{n})^{nt}-1]}{\frac{r}{n}}$ , where

1. $P_{t}$ is the balance in the account after t years

2. d is the regular deposit (the amount you deposit each year or month

or .........)

3. r is the annual interest rate in decimal

4. n is the number of compounding periods in one year

5. t the number of years

∵ Loren knows that he will have $500,000 when he retires

∴ $P_{t}$ = $500,000

∵ he sets up a payout annuity for 30 years in an account paying 10%

interest

∴ t = 30 years

∴ r = (10/100) = 0.1

∵ The annuity could provide each month

∴ n = 12

Substitute the values above in the formula

∴ $500,000=\frac{d[(1+\frac{0.1}{12})^{(12)(30)}-1]}{\frac{0.1}{12}}$

∴ $500,000=\frac{d[(1.00833333)^{360}-1]}{0.00833333}$

∴ 500,000 = d [2260.487925]

- Divide both sides by 2260.487925

∴ d = $221.19

The annuity could provide $221.19 each month

Learn more:

You can learn more about deposit in brainly.com/question/8782252

#LearnwithBrainly

8 0

3 years ago

Which weights are within 2 standard deviations of the mean? select three options. 8.4 lbs 8.9 lbs 9.5 lbs 10.4 lbs 10.9 lbs

djverab [1.8K]

The weights are within 2 standard deviations of the mean are 8.9 lbs, 9.5 lbs and 10.4 lbs

<h3>How to determine the weights?</h3>

The given parameters are:

Mean, μ = 9.5
Standard deviation, σ = 0.5

The weights within 2 standard deviation is represented as:

μ - 2σ ≤ x ≤ μ + 2σ

Substitute known values

9.5 - 2(0.5) ≤ x ≤ 9.5 + 2(0.5)

Evaluate the product

9.5 - 1 ≤ x ≤ 9.5 + 1

Evaluate the sum

8.5 ≤ x ≤ 10.5

This means that the weights are between 8.5 and 10.5 (inclusive)

Hence, the weights are within 2 standard deviations of the mean are 8.9 lbs, 9.5 lbs and 10.4 lbs

Read more about standard deviation at:

brainly.com/question/11743583

5 0

2 years ago