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ale4655 [162]
3 years ago
10

A pet store has 8 ​puppies, including 3 ​poodles, 2 ​terriers, and 3 retrievers. If Rebecka selects one puppy at​ random, the pe

t store replaces the puppy with a puppy of the same​ breed, then Aaron chooses a puppy at random. Find the probability that they both select a poodle.
Mathematics
1 answer:
Debora [2.8K]3 years ago
6 0

Answer:

9 / 64

Step-by-step explanation:

- In this task you have 2 events and you are looking for a joint probability. The first event is "Rebecca chooses a poodle". The probability of this event is:

                    P ( Rebecca chooses a poodle ) = 3 / 8

- because among 8 dogs there are 3 poodles.

- The second event is "Aaron selects a poodle".

This event has a probability of  that is equivalent to previous selection:

                   P ( Aaron chooses a poodle ) = 3 / 8

- Because after Rebecca's choice the chosen poodle is replaced with the poodle; hence, there are 8 pets in total and among them there are 3 poodles.

- To calculate probability of both events ("Rebeca selects a poodle and Aaron selects a poodle") with replacement you have to multiply both calculated probabilities - condition of independent events :

     P ( Aaron and Rebecca both select poodle ) = 3 / 8 * 3 / 8

                                                                                = 9 / 64

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Answer:

• No

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Step-by-step explanation:

To determine if the 4 given values of y are solutions to the inequality, start by solving the inequality. Solving an inequality is just like that of an equation, except that the direction of the sign changes when the inequality is divided by a negative number.

-2y +7≤ -5

Subtract 7 on both sides:

-2y≤ -5 -7

-2y≤ -12

Divide by -2 on both sides:

y≥ 6

This means that the solution can be 6 or greater than 6.

-10 and 3 are smaller than 6 and are not a solutions, while 7 and 6 satisfies the inequality and are therefore solutions.

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Alternatively, we can also substitute each value of y into the inequality and check if the value is less than or equal to -5.

Here's an example to check if -10 is a solution.

-2y +7≤ -5

When y= -10,

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= -2(-10) +7

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Since 27 is greater than 5, it is <u>not</u> a solution to the inequality.

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Suppose the test scores for a college entrance exam are normally distributed with a mean of 450 and a s. d. of 100. a. What is t
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Answer:

a) 68.26% probability that a student scores between 350 and 550

b) A score of 638(or higher).

c) The 60th percentile of test scores is 475.3.

d) The middle 30% of the test scores is between 411.5 and 488.5.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 450, \sigma = 100

a. What is the probability that a student scores between 350 and 550?

This is the pvalue of Z when X = 550 subtracted by the pvalue of Z when X = 350. So

X = 550

Z = \frac{X - \mu}{\sigma}

Z = \frac{550 - 450}{100}

Z = 1

Z = 1 has a pvalue of 0.8413

X = 350

Z = \frac{X - \mu}{\sigma}

Z = \frac{350 - 450}{100}

Z = -1

Z = -1 has a pvalue of 0.1587

0.8413 - 0.1587 = 0.6826

68.26% probability that a student scores between 350 and 550

b. If the upper 3% scholarship, what score must a student receive to get a scholarship?

100 - 3 = 97th percentile, which is X when Z has a pvalue of 0.97. So it is X when Z = 1.88

Z = \frac{X - \mu}{\sigma}

1.88 = \frac{X - 450}{100}

X - 450 = 1.88*100

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X when Z has a pvalue of 0.60. So it is X when Z = 0.253

Z = \frac{X - \mu}{\sigma}

0.253 = \frac{X - 450}{100}

X - 450 = 0.253*100

X = 475.3

The 60th percentile of test scores is 475.3.

d. Find the middle 30% of the test scores.

50 - (30/2) = 35th percentile

50 + (30/2) = 65th percentile.

35th percentile:

X when Z has a pvalue of 0.35. So X when Z = -0.385.

Z = \frac{X - \mu}{\sigma}

-0.385 = \frac{X - 450}{100}

X - 450 = -0.385*100

X = 411.5

65th percentile:

X when Z has a pvalue of 0.35. So X when Z = 0.385.

Z = \frac{X - \mu}{\sigma}

0.385 = \frac{X - 450}{100}

X - 450 = 0.385*100

X = 488.5

The middle 30% of the test scores is between 411.5 and 488.5.

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URGENT, HERES THE EQUATION!:)
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The answer to the question would be 3.5
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