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wlad13 [49]
2 years ago
15

Rahim was asked to make p the subject of this formula

Mathematics
1 answer:
KonstantinChe [14]2 years ago
4 0

Answer:

P =3Q +21

Step-by-step explanation:

Given: Rahim was asked to make p the subject of this formula

Q = p/3 - 7

He started his answer as follows

3Q = p - 21 hence p =

To Find, Complete Rahim's final answer.

Solution:

Q = p/3 - 7

Multiply both sides by 3

3Q = 3(p/3 - 7)

3Q = 3 × p/3 - 3 × 7

3Q = p - 21

Adding 21 both sides

3Q + 21 = p -21 +21

3Q + 21 = p

interchange sides

p = 3Q + 21

hence p =3Q +21

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Mkey [24]

Answer:

HI = 8

TU = 45

Step-by-step explanation:

→ Find the scale factor enlargement

45 ÷ 10 = 4.5

→ Multiply JI by the scale factor to find it

10 × 4.5 = 45

→ Divide UV by the scale factor to find it

36 ÷ 4.5 = 8

3 0
2 years ago
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Subtract. Write the difference in simplest form. (6p + 3) - (2p + 1)
Sonja [21]

Answer:

4p + 2

Step-by-step explanation:

(6p + 3) - (2p + 1)

Distribute the minus sign

6p +3 -2p -1

4p +2

3 0
3 years ago
Enter the simplified form of the complex fraction in the box.
Ket [755]

The problem is:

\frac{\frac{3}{x}+\frac{1}{4}}{1+\frac{3}{x}}

We now do L.C.M. of numerator and add, also do L.C.M. of denominator and add together:

\frac{\frac{12+x}{4x}}{\frac{x+3}{x}}

Dividing by a fraction is same as multiplying by its reciprocal, so we have:

\frac{12+x}{4x} * \frac{x}{x+3} \\

x cancels out, so can multiply and write:

\frac{12+x}{4} * \frac{1}{x+3} \\=\frac{12+x}{4x+12}

This is the simplified form.


ANSWER: \frac{12+x}{4x+12}

3 0
2 years ago
9. A large electronic office product contains 2000 electronic components. Assume that the probability that each component operat
Marysya12 [62]

Answer:

97.10% probability that five or more of the original 2000 components fail during the useful life of the product.

Step-by-step explanation:

For each component, there are only two possible outcomes. Either it works correctly, or it does not. The probability of a component falling is independent from other components. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

In this problem we have that:

n = 2000, p = 1-0.995 = 0.005

Approximate the probability that five or more of the original 2000 components fail during the useful life of the product.

We know that either less than five compoenents fail, or at least five do. The sum of the probabilities of these events is decimal 1. So

P(X < 5) + P(X \geq 5) = 1

We want P(X \geq 5)

So

P(X \geq 5) = 1 - P(X < 5)

In which

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{2000,0}.(0.005)^{0}.(0.995)^{2000} = 0.000044

P(X = 1) = C_{2000,1}.(0.005)^{1}.(0.995)^{1999} = 0.000445

P(X = 2) = C_{2000,2}.(0.005)^{2}.(0.995)^{1998} = 0.002235

P(X = 3) = C_{2000,3}.(0.005)^{3}.(0.995)^{1997} = 0.007480

P(X = 4) = C_{2000,4}.(0.005)^{4}.(0.995)^{1996} = 0.018765

P(X < 5) = P(X = 0) + `P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.000044 + 0.000445 + 0.002235 + 0.007480 + 0.018765 = 0.0290

P(X \geq 5) = 1 - P(X < 5) = 1 - 0.0290 = 0.9710

97.10% probability that five or more of the original 2000 components fail during the useful life of the product.

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3 years ago
The product of 2 and the second power of y
Morgarella [4.7K]
The answer is y2*2 

hope this helps
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