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Svet_ta [14]
3 years ago
9

There is a stack of 10 cards, each given a different number from 1 to 10. Suppose we select a card randomly from the stack, repl

ace it, and then randomly select another card. What is the probability that the first card is an even number and the card is greater than 7? Write your answer as a fraction in simplest form.
Mathematics
2 answers:
Greeley [361]3 years ago
8 0

Answer:

3/20

Step-by-step explanation:

There are 5 even numbers and 3 numbers greater than 7.

The probability of selecting an even number is 5/10 = 1/2

Since the card is replaced, the probability of a number greater than 7 is 3/10

So the probability of selecting an even card, then a card greater than 7 with replacement is 1/2 x 3/10 = 3/20

aev [14]3 years ago
4 0

Answer:

1/5 chance or 20% chance.

Step-by-step explanation:

When doing this problem we can start with only being able to go for 7 and up. This already makes us go down to a 3/10. Now we have 8, 9, and 10. Only two of those numbers are even 8, and 10. Which gives us a 2/10. 2/10 then simplifies to 1/5.

:D

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Meredith picked four times as many green peppers as red peppers if she picked a total of 20 peppers how many green peppers did s
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3 years ago
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FinnZ [79.3K]

Answer: a) -6, b) 32, c) -48, d) 9, e) -12

Step-by-step explanation:

Since we have given that

A and B are 4 × 4 matrices.

Here,

det (A) = -3

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We need to find the respective parts:

a) det (AB)

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c) det (2A)

Since we know that

\mid kA\mid =k^n\mid A\mid

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d) \bold{det(A^TA)}

Since we know that

\mid A^T\mid=\mid A\mid

so, it becomes,

\mid A^TA\mid=\mid A^T\mid \times \mid A\mid=-3\times -3=9

e) det (B⁻¹AB)

As we know that

\mid B^{-1}\mid =\mid B\mid

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\mid B^{-1}AB}\mid =\mid B^{-1}.\mid \mid A\mid.\mid B\mid=2\times -3\times 2=-12

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6 0
4 years ago
Sorry just one question please help me Desperate :'(
alex41 [277]
I think it’s the second option
7 0
4 years ago
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