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masya89 [10]
3 years ago
13

for the following two numbers, find two factors of the first number such that their product is the first number and their sum is

the second number ? -21,-4
Mathematics
1 answer:
zepelin [54]3 years ago
5 0
You can make some algebraic equations and solve it.
The first would be:
x \times y =  - 21
The second would be
x + y =  - 4
You can then rearrange the second into
x =  - 4 - y
And subsitute it into the first like so:
( - y - 4) \times y = -  21
After that, distribute the y into the parantheses.
{ - y}^{2}  - 4y =  - 21
Subtract the 21 on both sides and multiply by -1 on both sides:
{ - y}^{2}  - 4y  +  21 \\ {y}^{2}  + 4y  -  21
You then can factor it into:
(y + 3) \times (y - 7) = 0
With Zero Product Property, we can determine y to be either -3 and 7. Since the variables are interchangable, you can say the same about x, just that whatever x is, y must be the other value.


Thus, the answer is 7 and -3.
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A box contains 11 red chips and 4 blue chips. We perform the following two-step experiment: (1) First, a chip is selected at ran
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P(B1) = (11/15)

P(B2) = (4/15)

P(A) = (11/15)

P(B1|A) = (5/7)

P(B2|A) = (2/7)

Step-by-step explanation:

There are 11 red chips and 4 blue chips in a box. Two chips are selected one after the other at random and without replacement from the box.

B1 is the event that the chip removed from the box at the first step of the experiment is red.

B2 is the event that the chip removed from the box at the first step of the experiment is blue. A is the event that the chip selected from the box at the second step of the experiment is red.

Note that the probability of an event is the number of elements in that event divided by the Total number of elements in the sample space.

P(E) = n(E) ÷ n(S)

P(B1) = probability that the first chip selected is a red chip = (11/15)

P(B2) = probability that the first chip selected is a blue chip = (4/15)

P(A) = probability that the second chip selected is a red chip

P(A) = P(B1 n A) + P(B2 n A) (Since events B1 and B2 are mutually exclusive)

P(B1 n A) = (11/15) × (10/14) = (11/21)

P(B2 n A) = (4/15) × (11/14) = (22/105)

P(A) = (11/21) + (22/105) = (77/105) = (11/15)

P(B1|A) = probability that the first chip selected is a red chip given that the second chip selected is a red chip

The conditional probability, P(X|Y) is given mathematically as

P(X|Y) = P(X n Y) ÷ P(Y)

So, P(B1|A) = P(B1 n A) ÷ P(A)

P(B1 n A) = (11/15) × (10/14) = (11/21)

P(A) = (11/15)

P(B1|A) = (11/21) ÷ (11/15) = (15/21) = (5/7)

P(B2|A) = probability that the first chip selected is a blue chip given that the second chip selected is a red chip

P(B2|A) = P(B2 n A) ÷ P(A)

P(B2 n A) = (4/15) × (11/14) = (22/105)

P(A) = (11/15)

P(B2|A) = (22/105) ÷ (11/15) = (2/7)

Hope this Helps!!!

5 0
3 years ago
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