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Zolol [24]
4 years ago
7

What is the factoring?

Mathematics
1 answer:
Zigmanuir [339]4 years ago
6 0
I hope this helps you

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5 + 7 x 4 – (11 + 6)
satela [25.4K]

Answer:16

Step-by-step explanation:

Start with whats inside (11+6) (17) then you will multiply 7 and 4 = 28+5 =33 then back to 17,,,, 33-17=16

3 0
3 years ago
Read 2 more answers
What is 1 divided by 1/5
masha68 [24]
Your answer to this is:
0.2
3 0
4 years ago
Read 2 more answers
Help help help thanks
Dmitry [639]

Answer: a=2/5b + 1/5c (I may have answered your question but idk if I did)

Step 1: Add -4a to both sides.

9a−2b+−4a=4a+c+−4a

5a−2b=c

Step 2: Add 2b to both sides.

5a−2b+2b=c+2b

5a=2b+c

Step 3: Divide both sides by 5.

5a/5=2b+c/5

a=2/5b+1/5c

7 0
3 years ago
2 QUESTIONS LEFTT 20 POINTSS
AfilCa [17]

Answer:

C

Step-by-step explanation:

can I be brainliest?

8 0
3 years ago
According to the Center for Disease Control and Prevention (CDC), up to 20% of Americans contract the influenza virus each year,
12345 [234]

Answer:

(1) a. 0.0009

(2) d. 0.640

(3)

  • a. P(A and B) = 0.06.
  • b. P(A or B) = 0.70.

(4)Not disjoint

(5) a. nearly 0.

(6)b. 0.919

Step-by-Step Explanation:

(1)Probability of a baby being born with a birth defect =3%=0.03

The probability that both babies have birth defects=0.03 X 0.03= 0.0009.

(2)The probability of contracting the influenza virus each year = 20%=0.2

Therefore, the probability of not contracting the influenza virus =1-0.2=0.8

The probability that neither baby catches the flu in a given year:

=0.8 X  0.8

=0.64

(3)

P(A)=0.1

P(B)=0.6

P(A or B)=P(A)+P(B)=0.1 + 0.6 =0.7

P(A and B)=P(A)XP(B)=0.1 X 0.6 =0.06

(4)

P(A)=0.2

P(B)=0.9

Event A and B cannot be disjoint.

(5)

The probability of an American woman aged 20 to 24 having Chlamydia infection  =\dfrac{2791.5}{100000}

The probability that three randomly selected women in this age group have the infection

=\dfrac{2791.5}{100000} \times \dfrac{2791.5}{100000} \times \dfrac{2791.5}{100000} \\\\=0.00002175\\\approx 0

(6)The probability of an American woman aged 20 to 24 not having Chlamydia infection  =1-\dfrac{2791.5}{100000}

The probability that three randomly selected women in this age group do not have the infection

=\left(1-\dfrac{2791.5}{100000}\right)^3\\\\=0.9186\\\approx 0.919

7 0
4 years ago
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