Multiply 8-2x/5* 45x^3+90^2/10x^2-20x-80

you spent half of your weekly allowance at the movies. to earn more money your parents let you wash the car for $8. what is your

Brrunno [24]

20. 1/2 (n) + 8 = $20

4 0

3 years ago

Suppose the mean GPA of all students graduating from a particular university in 1975 was 2.40. The registrar plans to look at re

ycow [4]

Answer:

Correct option:

(B) H₀: μ = 2.40 vs. Hₐ: μ ≠ 2.40.

Step-by-step explanation:

The registrar of particular university in 1975 plans to look at records of students graduating last year to see if the mean GPA has changed from 2.40.

The registrar can use a single mean test to determine whether the mean has changed or not.

The hypothesis can be described as:

H₀: The mean GPA is 2.40, i.e. μ = 2.40.

Hₐ: The mean GPA is different from 2.40, i.e. μ ≠ 2.40.

To perform the test the registrar can either use a z-distribution or a t-distribution.

If the data provided gives some insight about the population standard deviation and the sample selected is quite large then the z-distribution can be used.

Otherwise it is wiser to use a t-distribution.

The decision rule is:

If the p-value of the test is less than the significance level then the null hypothesis will be rejected and vice-versa.

Thus, the correct option is (B).

6 0

3 years ago

I need help asap first answer gets brainly

LuckyWell [14K]

Hey there! I'm happy to help!

The domain is any possible number you can input into the function to get a real output. The domain of h just means the domain of this entire function, which is called h.

Let's look at the answer options.

OPTION A

All real values of x such that x≠0.

The only way to make it so that we do not have a real output is if we get a negative square root. You cannot multiply any number by itself to get a negative number unless you use imaginary numbers, but using imaginary numbers makes our output not real.

Anyways, plugging in 0 would give us √-10, which is not a real number. That part is correct, but this option says ALL REAL NUMBERS except for 0. The problem is is that we can take any number less than ten and plug it in and we would get a negative square root, a fake number. So, this option is incorrect.

OPTION B

All real values of x such that x≥10.

Let's say we use 10 for our x and plug it in. This gives us √0, which is 0, a real output. Anything bigger than this 10 will give us a real output as well, so this option is correct.

We don't even need to check the other options because we have already found the correct answer. C,D, and E are all incorrect though because they include values less ten, which would give us a negative square root, a fake number.

I hope that this helps! Have a wonderful day!

8 0

3 years ago

Miguel is making a snack mix. The recipe calls for six parts of spicy tortilla chips to three parts of corn chips. Miguel would

Harlamova29_29 [7]

Equation:
6x + 3x = 45

To find X:
6x = 45
45/6 = 7.5
x = 7.5

To find other X:
3x = 45
45/3 = 15
x = 15

Final answer:

7.5 cups tortilla chips

15 cups corn chips

3 0

3 years ago

Read 2 more answers

Find the absolute extrema of the function over the region R. (In each case, R contains the boundaries.) Use a computer algebra s

algol [13]

f(x, y) = x ² - 4xy + 5

has critical points where both partial derivatives vanish:

∂f/∂x = 2x - 4y = 0 ==> x = 2y

∂f/∂y = -4x = 0 ==> x = 0 ==> y = 0

The origin does not lie in the region R, so we can ignore this point.

Now check the boundaries:

• x = 1 ==> f (1, y) = 6 - 4y

Then

max{f (1, y) | 0 ≤ y ≤ 2} = 6 when y = 0

max{f (1, y) | 0 ≤ y ≤ 2} = -2 when y = 2

• x = 4 ==> f (4, y) = 12 - 16y

Then

max{f (4, y) | 0 ≤ y ≤ 2} = 12 when y = 0

max{f (4, y) | 0 ≤ y ≤ 2} = -4 when y = 2

• y = 0 ==> f (x, 0) = x ² + 5

Then

max{f (x, 0) | 1 ≤ x ≤ 4} = 21 when x = 4

min{f (x, 0) | 1 ≤ x ≤ 4} = 6 when x = 1

• y = 2 ==> f (x, 2) = x ² - 8x + 5 = (x - 4)² - 11

Then

max{f (x, 2) | 1 ≤ x ≤ 4} = -2 when x = 1

min{f (x, 2) | 1 ≤ x ≤ 4} = -11 when x = 4

So to summarize, we found

max{f(x, y) | 1 ≤ x ≤ 4, 0 ≤ y ≤ 2} = 21 at (x, y) = (4, 0)

min{f(x, y) | 1 ≤ x ≤ 4, 0 ≤ y ≤ 2} = -11 at (x, y) = (4, 2)

5 0

3 years ago