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viktelen [127]
3 years ago
5

The alphabet game costs $.25 to play. Before the game, 26 slips of paper with a different letter of the alphabet on it are put i

nto a bag. A player draws one slip from the bag. If the player draws a vowel (A, E, I, O, or U), he or she wins $1. If a player plays the alphabet game 2 times in a row, replacing the slip of paper after each turn, what is the probability that they win twice? Write as a fraction.
Mathematics
1 answer:
iragen [17]3 years ago
3 0

Answer: 25/676

Step-by-step explanation:

Number of possible outcomes = 26

In other to win, one must draw must be either (A, E, I, O or U)

Therefore required drws to win = 5

First draw:

P(win) = Total required outcome / Total possible outcome

P(win) = 5/26

Second draw:

P(win) = Total required outcome / Total possible outcome

P(win) = 5/26

Therefore,

P(winning twice) = (5/26) × (5/26) = 25/676

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Answer:

1 week

Step-by-step explanation:

<u><em>The correct question is</em></u>

Amira has 3/4 of a bag of cat food .Her cat eats 3/4 of a bag per week . How many weeks will the food last ?

we know that

To determine how many weeks her food will last, we take the amount of food and divide by how much she feeds her cats.

3/4 bag ÷ 3/4 bag per week

\frac{3}{4}: \frac{3}{4}

Multiply in cross

\frac{3*4}{4*3}=1\ week

8 0
3 years ago
Swimming Fleury swam 6 laps the first day,15 laps the second day,24 laps the third day, and so on.On which day will fleury swim
mihalych1998 [28]
Fleury will swim 51 laps on the sixth day if this pattern continues.
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Solving right triangles
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3 years ago
A research company desires to know the mean consumption of meat per week among people over age 40. A sample of 610 people over a
Papessa [141]

Answer: (3.0,\ 3.2)

Step-by-step explanation:

Confidence interval for population mean is given by :-

\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}  (1)

, where \overline{x} = Sample mean

z_{\alpha/2}= Critical z-value

\sigma = Population standard deviation.

n= Sample size.

As per given , we have

n= 610

\sigma=1.1

\overline{x}=3.1

Significance level for 85% confidence : \alpha=1-0.85=0.15

By z-table critical two tailed z-value : z_{\alpha/2}=z_{0.075}=1.44

Put all values in (1) , we get

3.1\pm 1.44\dfrac{1.1}{\sqrt{610}}

3.1\pm 1.44\dfrac{1.1}{24.698}

3.1\pm 0.064

=(3.1-0.064,\ 3.1+0.064)=(3.036,\ 3.164)\approx(3.0,\ 3.2)

Hence, the 85% confidence interval for the mean consumption of meat among people over age 40.  = (3.0,\ 3.2)

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We have been provided a diagram which tells us that Patti drew vertical line segments from two points to the line in her scatter plot. The first point she selected was dwarf crocodile. The second point she selected was for an Indian Gharial crocodile.

We can see that dwarf crocodile's bite force is closer to line of best fit than Indian Gharial crocodile. Indian Gharial crocodile seems to be an outlier for our data set.

Therefore, Patti's line have resulted in a predicted bite force that was closer to actual bite force for the dwarf crocodile.

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