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Sever21 [200]
3 years ago
15

Pleases show workkkkkkkkkkkkkkkk

Mathematics
1 answer:
melomori [17]3 years ago
7 0

Answer:

5) 2(4 - n) = -3n

2 × 4 - 2 × n = -3n

8 - 2n = -3n

-2n + 3n = -8

1n = -8

n = -8/1

n = -8

6) 5 - 3(2 + 2w) = -7

5 - 3 × 2 + -3 × 2w = -7

5 - 6 - 6w = -7

-1 - 6w = -7

-6w = -7 + 1

-6w = -6

w = -6/-6

w = 1

7) 5(2r + 3) - 5 = 0

5 × 2r + 5 × 3 - 5 = 0

10r + 15 - 5 = 0

10r + 10 = 0

10r = -10

r = -10/10

r = -1

8) 3 - 5(2d - 3) = 4

3 - 5 × 2d - 5 × -3 = 4

3 - 10d + 15 = 4

-10d + 15 + 3 = 4

-10d + 18 = 4

-10d = 4 - 18

-10d = -14

d = -14/-10

d = 7/5

hope this helps you!!

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Step-by-step explanation:

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3 years ago
What is it 7 points on the board for u plz answer CORRECTTT
pochemuha

Answer:

x is 30

Step-by-step explanation:

1/2 ( x + 6 ) = 18

1/2x + 3 = 18

       - 3    - 3

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What is the partial product of 434x310
Svetach [21]
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4 years ago
Read 2 more answers
Geometry help please.
m_a_m_a [10]
The answer is D.) (4,3) 

trace the line to where the middle of the point looks to be and then find the point. in this instance the segment is 12 points long so the middle would be at six but because it is shifted over to the left 2 points the x coordinate would be 4. Since the line is parallel with the x axis you know that the y coordinate has to be 3. So the answer is (4,3)
4 0
4 years ago
A company is interviewing potential employees. Suppose that each candidate is either qualified, or unqualified with given probab
stira [4]

Answer:

P(C=1|T=1)=q(\sum_{i=15}^{20}\binom{20}{i} p^i(1-p)^{20-i})( \sum_{i=15}^{20}\binom{20}{i}[qp^i(1-p)^{20-i} + (1-q)p^{20-i}(1-p)^i])^{-1}

Step-by-step explanation:

Hi!

Lets define:

C = 1  if candidate is qualified

C = 0 if candidate is not qualified

A = 1 correct answer

A = 0 wrong answer

T = 1 test passed

T = 0 test failed

We know that:

P(C=1)=q\\P(A=1 | C=1) = p\\P(A=0 | C=0) = p

The test consist of 20 questions. The answers are indpendent, then the number of correct answers X has a binomial distribution (conditional on the candidate qualification):

P(X=x | C=1)=f_1(x)=\binom{20}{x}p^x(1-p)^{20-x}\\P(X=x | C=0)=f_0(x)=\binom{20}{x}(1-p)^xp^{20-x}

The probability of at least 15 (P(T=1))correct answers is:

P(X\geq 15|C=1)=\sum_{i=15}^{20}f_1(i)\\P(X\geq 15|C=0)=\sum_{i=15}^{20}f_0(i)\\

We need to calculate the conditional probabiliy P(C=1 |T=1). We use Bayes theorem:

P(C=1|T=1)=\frac{P(T=1|C=1)P(C=1)}{P(T=1)}\\P(T=1) = qP(T=1|C=1) + (1-q)P(T=1|C=0)

P(T=1)=q\sum_{i=15}^{20}f_1(i) + (1-q)\sum_{i=15}^{20}f_0(i)\\P(T=1)=\sum_{i=15}^{20}\binom{20}{i}[qp^i(1-p)^{20-i} + (1-q)p^{20-i}(1-p)^i)]

P(C=1|T=1)=q(\sum_{i=15}^{20}\binom{20}{i} p^i(1-p)^{20-i})( \sum_{i=15}^{20}\binom{20}{i}[qp^i(1-p)^{20-i} + (1-q)p^{20-i}(1-p)^i])^{-1}

5 0
3 years ago
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