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Vinvika [58]
3 years ago
9

the sum of the first twelve terms of arithmetic progression [AP] is 168. if the third term is 7. find the value of the common di

fference and the first term ​
Mathematics
1 answer:
34kurt3 years ago
3 0

Answer:

Common difference = 2

First term = 3.

Step-by-step explanation:

3rd term = a1 + (3 - 1)d  = 7    where a1 = first term and d = the common difference.

Sum of the first 12 terms = (12/2)[2a1 + (12-1)d] = 168

so 6(2a1 + 11d) = 168

So simplifying, we have the following system:

 a1 + 2d = 7  ..................(A)

12a1 + 66d = 168

2a1 + 11d = 28.................(B)   Multiplying equation A by - 2:

-2a1 - 4d =- -14                      Adding this to equation B

7d = 14

d = 2

Now plug this into equation A:

a1 + 2(2) = 7

a1 = 7-4

a1 = 3.

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Answer:

B. Yes, because P(X ≥ 68) ≤ .05

Step-by-step explanation:

The probability of a score X occuring is said to be unusually high if the probability of a score of X or larger is lower than 5%.

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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.

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And p is the probability of X happening.

In this problem we have that:

Three out of four students said that courts show too much concern for criminals. This means that p = \frac{3}{4} = 0.75

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We have to find P(X \geq 68).

P(X \geq 68) = P(X = 68) + P(X = 69) + P(X = 70)

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P(X = 68) = C_{70,68}.(0.75)^{68}.(0.25)^{2} < 0.000001

P(X = 69) = C_{70,69}.(0.75)^{69}.(0.25)^{1} < 0.000001

P(X = 70) = C_{70,70}.(0.75)^{70}.(0.25)^{0} < 0.000001

So

P(X \geq 68) = P(X = 68) + P(X = 69) + P(X = 70) < 0.000003

P(X \geq 68) is lower than 0.05, so this is an unusually high number.

So the correct answer is:

B. Yes, because P(X ≥ 68) ≤ .05

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