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3 years ago

6

points m and n are two vertices of a right triangle. they are the endpoints of the hypotenuse mn of the triangle. the third vert

ex lies in the fourth quadrant

1 answer:

Alekssandra [29.7K]3 years ago

4 0

Answer:

HEY IMAGE IS BLUR

Step-by-step explanation:

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emma gave 2/5 of her money and an additional $9 to olivia. she gave 4/7 of her remaining money and an additional $1 to Ava. emma

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Yo wsp how is it going kane

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3 years ago

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Austin purchase 2/3 of a pound of yum yum treats. If yum yum treats are $6.00 per pound how much does Austin pay

Oksi-84 [34.3K]

1. The information given in the problem above, is:

- Austin purchases <span>2/3 of a pound of yum yum treats.
- Y</span><span>um yum treats are $6.00 per pound (1 pound=$6.00).

2. Keeping this on mind you have: If 1 pound of </span>yum yum treats is $6.00, how much is 2/3 of a pound?

1 pound------------$6.00
2/3 of a pound----x

x=(2/3)(6.00)/1
x=(2/3)(6.00)
x=12/3
x=$4.00
<span>
Therefore: How much does Austin pay?
</span>
The answer is: Austin pays $4.00.

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3 years ago

CAN SOMEONE HELP ME WITH THIS QUESTION!!!

earnstyle [38]

Answer:

E

Step-by-step explanation:

let a, b, c represent the number of students in 6th, 7th, 8th grade

ratio of students : teachers = 28 : 1

There are 82 teachers , so 28 × 82 = 2296 students

Then

a + b + c = 2296 , that is

828 + b + c = 2296 ( subtract 828 from both sides )

b + c = 1468 → E

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3 years ago

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kirill115 [55]

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3 years ago

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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 <em>sampling distribution of the sampling proportions</em>.

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.

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3 years ago