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andrey2020 [161]
3 years ago
10

in a standard deck of t2 playing cards, there are 4 aces, 4 kings, 4 queens, and 4 Jacks. All other cards have number values. th

e kings, queens, and Jack's are called face cards. You are dealt 2 cards in a sequence. calculate the probability of being dealt a 3 and then a 10. write your answer as a decimal to the nearest hundredth​
Mathematics
1 answer:
Illusion [34]3 years ago
3 0

Answer:

0.01

Step-by-step explanation:

Assuming the deck is the national playing deck with 52 cards.

Your probability of getting a 3 from the entire deck is 4/52. This number is the same for 10.

To get the probability of pulling a 3 and a 10 afterwards, you must multiply these two fractions (you're expecting a lower percentage).

4/52 * 4/52 = 0.0059

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Nimfa-mama [501]

I think the question was partially cut off, but I think the question was "Simplify (3x)^-1".

The answer to that is 1/3x, which is (B).

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5 pencils to 2 pens is 5/2 in simplest form
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Someone please do this problem for me.​
hoa [83]

Given:

$\frac{6}{a^{2}-7 a+6}, \frac{3}{a^{2}-36}

To find:

The LCD of the fractions.

Solution:

LCD means least common denominator.

Let us find the least common multiplier for the denominator.

The denominators are \left(a^{2}-7 a+a\right),\left(a^{2}-36\right).

Factor \left(a^{2}-7 a+a\right):

a^{2}-7 a+6=\left(a^{2}-a\right)+(-6 a+6)

Take a common in first 2 terms and -6 common in next two terms.

                  =a(a-1)-6(a-1)

Take out common factor (a - 1).

\left(a^{2}-7 a+a\right)=(a-1)(a-6) ------------- (1)

Factor \left(a^{2}-36\right):

\left(a^{2}-36\right)=\left(a^{2}-6^2\right)

Using identity: (a^2-b^2)=(a-b)(a+b)

\left(a^{2}-36\right)=(a-6)(a+6)  ------------- (1)

From (1) and (2),

LCM of \left(a^{2}-7 a+a\right),\left(a^{2}-36\right)=(a-1)(a-6)(a+6)

Therefore LCD is (a - 1)(a - 6)(a + 6).

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3 years ago
Suppose has a solution. Explain why the solution is unique precisely when has only the trivial solution. Choose the correct answ
Margarita [4]

Complete question is;

Suppose Ax = b has a solution. Explain why the solution is unique precisely when Ax = 0 has only the trivial solution. Choose the correct answer.

A. Since Ax = b is inconsistent, its solution set is obtained by translating the solution set of Ax = 0. For Ax = b to be inconsistent, Ax = 0 has only the trivial solution.

B. Since Ax = b is consistent, its solution set is obtained by translating the solution set of Ax = 0. So the solution set of Ax = b is a single vector if and only if the solution set of Ax = 0 is a single vector, and that happens if and only if Ax = 0 has only the trivial solution.

C. Since Ax = b is inconsistent, then the solution set of Ax = 0 is also inconsistent. The solution set of Ax = 0 is inconsistent if and only if Ax = 0 has only the trivial solution.

D. Since Ax = b is consistent, then the solution is unique if and only if there is at least one free variable in the corresponding system of equations. This happens if and only if the equation Ax = 0 has only the trivial solution.

Answer:

Option B: Since Ax = b is consistent, its solution set is obtained by translating the solution set of Ax = 0. So the solution set of Ax = b is a single vector if and only if the solution set of Ax = 0 is a single vector, and that happens if and only if Ax = 0 has only the trivial solution

Step-by-step explanation:

There are different ways of explaining this but we will explain it algebraic ally.

If Ax = b has a solution, then it can be said to be unique if and only if every column of A will be a pivot column. Now, If every column of A will be a pivot column, then it means that there are no free variables, and thus the homogeneous equation will have only the trivial solution.

Also homogeneous equations are always constant.

Thus the correct option is Option B: Since Ax = b is consistent, its solution set is obtained by translating the solution set of Ax = 0. So the solution set of Ax = b is a single vector if and only if the solution set of Ax = 0 is a single vector, and that happens if and only if Ax = 0 has only the trivial solution

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