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3 years ago

Ben makes $39,600 a year. What is the maximum amount he can afford for a'

Mathematics

1 answer:

Olin [163]3 years ago

3 0

Answer:

B:825

Step-by-step explanation:

Hopes this helps:)

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ETHANHUNT HELP PLS The marked sides have equal lengths. What is the value of x?

dezoksy [38]

Here, both sides are equal, represented by - sign.
So, 4x + 12 = 60
4x = 60 - 12
x = 48/4
x = 12

In short, Your Answer would be 12

Hope this helps!

3 0

2 years ago

There are not asymptotes for the parent function f (x) =1/x True False

Georgia [21]

Answer:

True

Step-by-step explanation:

the parent function f(x) = 1/x has a vertical asymptote at x = 0 and a horizontal asymptote at y = 0

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3 years ago

BRAINLIEST! :3 How much is a $15 tip with the meal costing $18?

VARVARA [1.3K]

With the meal costing $18 and the tip being $15 the total will be $33

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3 years ago

Plzzz helpppppp nowwwww

snow_lady [41]

Answer:

x= 10

I hope i helped, best of luck!

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2 years ago

Read 2 more answers

There are two games involving flipping a coin. In the first game you win a prize if you can throw between 45% and 55% heads. In

Nina [5.8K]

Answer:

d) 300 times for the first game and 30 times for the second

Step-by-step explanation:

We start by noting that the coin is fair and the flip of a coin has a probability of 0.5 of getting heads.

As the coin is flipped more than one time and calculated the proportion, we have to use the sampling distribution of the sampling proportions.

The mean and standard deviation of this sampling distribution is:

$\mu_p=p\\\\ \sigma_p=\sqrt{\dfrac{p(1-p)}{N}}$

We will perform an analyisis for the first game, where we win the game if the proportion is between 45% and 55%.

The probability of getting a proportion within this interval can be calculated as:

$P(0.45$

referring the z values to the z-score of the standard normal distirbution.

We can calculate this values of z as:

$z_H=\dfrac{p_H-\mu_p}{\sigma_p}=\dfrac{(p_H-p)}{\sqrt{\dfrac{p(1-p)}{N}}}=\sqrt{\dfrac{N}{p(1-p)}}*(p_H-p)>0\\\\\\z_L=\dfrac{p_L-\mu_p}{\sigma_p}=\dfrac{p_L-p}{\sqrt{\dfrac{p(1-p)}{N}}}=\sqrt{\dfrac{N}{p(1-p)}}*(p_L-p)$

If we take into account the z values, we notice that the interval increases with the number of trials, and so does the probability of getting a value within this interval.

With this information, our chances of winning increase with the number of trials. We prefer for this game the option of 300 games.

For the second game, we win if we get a proportion over 80%.

The probability of winning is:

$P(p>0.8)=P(z>z^*)$

The z value is calculated as before:

$z^*=\dfrac{p^*-\mu_p}{\sigma_p}=\dfrac{p^*-p}{\sqrt{\dfrac{p(1-p)}{N}}}=\sqrt{\dfrac{N}{p(1-p)}}*(p^*-p)>0$

As (p*-p)=0.8-0.5=0.3>0, the value z* increase with the number of trials (N).

If our chances of winnings depend on P(z>z*), they become lower as z* increases.

Then, we can conclude that our chances of winning decrease with the increase of the number of trials.

We prefer the option of 30 trials for this game.

8 0

2 years ago