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FromTheMoon [43]

3 years ago

5

It is 7 times as much to go to an amusement park as it is to go to the movies if it is $9 to go to the movies how much is it to

go to the amusement park?

2 answers:

julsineya [31]3 years ago

3 0

It will be 63 dollars to go to a amusement park

Lemur [1.5K]3 years ago

3 0

It equals 63 because you have to times 9 and 7 to get the right answer

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Which is equivalent to 3sqrt8^x ?

creativ13 [48]

The correct answer is A.

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The Hudson high school wrestling team just won the state tournament and have been awarded a triangular pennant to hang on the wa

Eduardwww [97]

Answer:

3 feet.

Step-by-step explanation:

Base of the triangular pennant = 1.5 feet long.
Area of the triangular pennant = 2.25 Square feet.

Now, we know that:

Area of a Triangle $=\frac{1}{2}$ X Base X Height$

Substituting the given values, we have:

$2.25=\dfrac{1}{2}$ X 1.5 X Height\\\\2.25=$\dfrac{1.5}{2}$ X Height\\\\Cross multiply\\\\2.25*2=1.5 X Height\\\\Divide both sides by 1.5 to solve for the height\\\\\dfrac{2.25*2}{1.5}=\dfrac{1.5 X Height}{1.5}\\\\$Height of the Triangle =3 feet$

Therefore, the height of the pennant is 3 feet.

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3 years ago

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DENIUS [597]

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3 years ago

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consider the following scenario: Assume daily protein intake values in a population of athletes are known to have a standard dev

Kipish [7]

Answer:

The 95% confidence interval for the population mean daily protein intake is between 69.97g and 84.03g.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.025 = 0.975$ , so $z = 1.96$

Now, find M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.96*\frac{58.6}{\sqrt{267}} = 7.03$

The lower end of the interval is the sample mean subtracted by M. So it is 77 - 7.03 = 69.97g.

The upper end of the interval is the sample mean added to M. So it is 77 + 7.03 = 84.03g.

The 95% confidence interval for the population mean daily protein intake is between 69.97g and 84.03g.

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3 years ago