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1 year ago

PLEASE HELP THIS IS A TEST IT NEEDS TO BE DONE FAST AND CORRECTLY!

Mathematics

1 answer:

frutty [35]1 year ago

7 0

Answer:

Step-by-step explanation:

(-7,1)

I hope this helped <3

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Answer:

0.6

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72 in

Step-by-step explanation:

Formula: A=a+b/2 h

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For circle O, m CD=125 and m

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Step-by-step explanation:

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3 years ago

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A grocery wants to make two kinds of coffee. One consoles for $1.40 pound, and the other cells for $2.95 a pound. He wants to ma

steposvetlana [31]

Step-by-step explanation:

a lot of typos in the problem description.

I assume the price of the mix is 1:1 related to the prices of the individual types of coffee.

so, from what I understood we can define

x = pounds of the first type of coffee.

y = pounds of the second type of coffee.

x + y = 21 pounds

1.4x + 2.95y = 2.6(x + y) = 2.6×21 = $54.60

in the second equation we have to bring the given "$2.60 per pound" to the total price, which is $2.60 times the total amount of pounds (which is 21).

from the first equation we get

x = 21 - y

and that we use in the second equation

1.4×(21 - y) + 2.95y = 54.6

29.4 - 1.4y + 2.95y = 54.6

1.55y = 25.2

y = 25.2 / 1.55 = 16.25806452... ≈ 16.26 pounds

x = 21 - y = 4.741935484... ≈ 4.74 pounds

so, he should mix 4.74 pounds of the first (cheaper) type and 16.26 pounds of the second (expensive) type.

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2 years ago

The compressive strength of concrete is normally distributed with mu = 2500 psi and sigma = 50 psi. A random sample of n = 8 spe

Nata [24]

Answer:

$X \sim N(2500,50)$

Where $\mu=2500$ and $\sigma=50$

We select a sample size of n =8. Since the distribution for X is normal then we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And the standard error would be:

$SE = \frac{50}{\sqrt{8}}= 17.678 psi$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

Let X the random variable that represent the compressive strength of concrete of a population, and for this case we know the distribution for X is given by:

$X \sim N(2500,50)$

Where $\mu=2500$ and $\sigma=50$

We select a sample size of n =8. Since the distribution for X is normal then we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And the standard error would be:

$SE = \frac{50}{\sqrt{8}}= 17.678 psi$

7 0

3 years ago