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love history [14]
3 years ago
10

The overhead reach distances of adult females are normally distributed with a mean of 202.5 cm and a standard deviation of

Mathematics
1 answer:
bezimeni [28]3 years ago
5 0

Answer:

C. The normal distribution can be used because the original population has a normal distribution.

Step-by-step explanation:

When the original population has a normal distribution, we can use the normal distribution to solve the exercise.

In this problem, we have that:

The original population are the adult females.

The problem states that:

The overhead reach distances of adult females are normally distributed with a mean of 202.5 cm and a standard deviation of

7.8 cm.

So the correct answer is

C. The normal distribution can be used because the original population has a normal distribution.

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Answer: No, you cannot.

Step-by-step explanation: A square number cannot be a perfect number.

The numbers: 0, 1, 5, 14, 30, 55, 91, 140, 204, 285, 385, 506, 650, 819, 1015, 1240, 1496, 1785, 2109, 2470, 2870, 3311, 3795, 4324, 4900, 5525, 6201 are examples of perfect numbers.

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Which unit of measure would be appropriate for the volume of a sphere with a radius of 2 meters?
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The volume of a figure is the quantity of three-dimensional space enclosed by a closed surface, or simpler the number of cubes required to fill it completely.

The basic unit of volume in the metric system is the liter (l).There are 1000 liters per cubic meter.

Howeever, the unit of measure that would be appropriate for the volume of a sphere with a radius of 2 meters is cubic meters. Correct answer: C


4 0
3 years ago
Mrs. Rodriguez is going to use 6 1/3 yards of material to make two dresses . the large dress requires3 2/3 yards of material. ho
Dmitry_Shevchenko [17]

Answer:

She has 6 1/3. The larger dress need 3 2/3. So 6 1/3 - 3 2/3 = 2 2/3

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Step-by-step explanation:

7 0
3 years ago
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yawa3891 [41]

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Does the answer help you?

please mark this question as brainliest answer.Cause it take a few mins to solve your question.Tq

5 0
3 years ago
Your friend asks if you would like to play a game of chance that uses a deck of cards and costs $1 to play. They say that if you
gtnhenbr [62]

Answer:

Expected value = 40/26 = 1.54 approximately

The player expects to win on average about $1.54 per game.

The positive expected value means it's a good idea to play the game.

============================================================

Further Explanation:

Let's label the three scenarios like so

  • scenario A: selecting a black card
  • scenario B: selecting a red card that is less than 5
  • scenario C: selecting anything that doesn't fit with the previous scenarios

The probability of scenario A happening is 1/2 because half the cards are black. Or you can notice that there are 26 black cards (13 spade + 13 club) out of 52 total, so 26/52 = 1/2. The net pay off for scenario A is 2-1 = 1 dollar because we have to account for the price to play the game.

-----------------

Now onto scenario B.

The cards that are less than five are: {A, 2, 3, 4}. I'm considering aces to be smaller than 2. There are 2 sets of these values to account for the two red suits (hearts and diamonds), meaning there are 4*2 = 8 such cards out of 52 total. Then note that 8/52 = 2/13. The probability of winning $10 is 2/13. Though the net pay off here is 10-1 = 9 dollars to account for the cost to play the game.

So far the fractions we found for scenarios A and B were: 1/2 and 2/13

Let's get each fraction to the same denominator

  • 1/2 = 13/26
  • 2/13 = 4/26

Then add them up

13/26 + 4/26 = 17/26

Next, subtract the value from 1

1 - (17/26) = 26/26 - 17/26 = 9/26

The fraction 9/26 represents the chances of getting anything other than scenario A or scenario B. The net pay off here is -1 to indicate you lose one dollar.

-----------------------------------

Here's a table to organize everything so far

\begin{array}{|c|c|c|}\cline{1-3}\text{Scenario} & \text{Probability} & \text{Net Payoff}\\ \cline{1-3}\text{A} & 1/2 & 1\\ \cline{1-3}\text{B} & 2/13 & 9\\ \cline{1-3}\text{C} & 9/26 & -1\\ \cline{1-3}\end{array}

What we do from here is multiply each probability with the corresponding net payoff. I'll write the results in the fourth column as shown below

\begin{array}{|c|c|c|c|}\cline{1-4}\text{Scenario} & \text{Probability} & \text{Net Payoff} & \text{Probability * Payoff}\\ \cline{1-4}\text{A} & 1/2 & 1 & 1/2\\ \cline{1-4}\text{B} & 2/13 & 9 & 18/13\\ \cline{1-4}\text{C} & 9/26 & -1 & -9/26\\ \cline{1-4}\end{array}

Then we add up the results of that fourth column to compute the expected value.

(1/2) + (18/13) + (-9/26)

13/26 + 36/26 - 9/26

(13+36-9)/26

40/26

1.538 approximately

This value rounds to 1.54

The expected value for the player is 1.54 which means they expect to win, on average, about $1.54 per game.

Therefore, this game is tilted in favor of the player and it's a good decision to play the game.

If the expected value was negative, then the player would lose money on average and the game wouldn't be a good idea to play (though the card dealer would be happy).

Having an expected value of 0 would indicate a mathematically fair game, as no side gains money nor do they lose money on average.

7 0
2 years ago
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