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user100 [1]
2 years ago
13

Emerson purchased a game that was on sale for 18% off. The sales tax in his county is 8%. Let y represent the original price of

the game. Write an expression that can be used to determine the final cost of the game.
(I will give 20 points but it only says 10!!!)
Mathematics
2 answers:
leva [86]2 years ago
6 0

Answer:

y(.82)*1.08

Step-by-step explanation:

You multiply by .82 to get the cost of the game after the discount and then by 1.08 to add the tax and total amount.  

marissa [1.9K]2 years ago
6 0

Answer: 1.08 * 0.82y

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Rational number 3 by 40 is equals to​
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A bin of 5 transistors is known to contain 2 that are defective. The transistors are to be tested, one at a time, until the defe
xxMikexx [17]

Answer:

P(N_1 = a , N_2 = b)= \frac{1}{5-a C 1} * \frac{5-a C 1}{5C2} = \frac{1}{5C2}=\frac{1}{10}

Step-by-step explanation:

For the random variable N_1 we define the possible values for this variable on this case [1,2,3,4,5] . We know that we have 2 defective transistors so then we have 5C2 (where C means combinatory) ways to select or permute the transistors in order to detect the first defective:

5C2 = \frac{5!}{2! (5-2)!}= \frac{5*4*3!}{2! 3!}= \frac{5*4}{2*1}=10

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P(N_1 = a) = \frac{5-a C 1}{5C2}

For the distribution of N_2 we need to take in count that we are finding a conditional distribution. N_2 given N_1 =a, for this case we see that N_2 \in [1,2,...,5-a], so then exist 5-a C 1 ways to reorder the remaining transistors. And if we want b additional steps to obtain a second defective transistor we have the following probability defined:

P(N_2 =b | N_1 = a) = \frac{1}{5-a C 1}

And if we want to find the joint probability we just need to do this:

P(N_1 = a , N_2 = b) = P(N_2 = b | N_1 = a) P(N_1 =a)

And if we multiply the probabilities founded we got:

P(N_1 = a , N_2 = b)= \frac{1}{5-a C 1} * \frac{5-a C 1}{5C2} = \frac{1}{5C2}=\frac{1}{10}

8 0
2 years ago
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Answer:

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Explanation: 

Since 5% of the values are missing, the expected number of records to be removed is interpreted as the expected value of the records to be removed whose probability of having missing values is 0.05. 

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