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bixtya [17]
2 years ago
9

What is the eleventh term in the sequence 17, 24, 31, 38…?

Mathematics
2 answers:
shusha [124]2 years ago
6 0

Answer:

87

Explanation:

This is an arithmetic sequence with <u>common difference</u>: 7 and <u>first term</u>: 17

Arithmetic sequence:

a + (n - 1)d                    where n is term position, d is difference, a is first term

17 + (n - 1)7

7n + 10

The eleventh term:

7n + 10

7(11) + 10

77 + 10

87

Ymorist [56]2 years ago
6 0

<u>To Find :</u><u>-</u>

  • Eleventh term of the sequence.

<u>Solution</u><u> </u><u>:</u><u>-</u>

Given A.P.,

  • 17 , 24 , 31 , 38 …

First term (a) is 17.

Common difference (d) = 24 - 17 => 7

We know that :

  • tn = a + (n - 1) d

Here n would be 11.

>> t11 = 17 + (11 - 1) 7

>> t11 = 17 + (10) 7

>> t11 = 17 + (10) × 7

>> t11 = 17 + 70

>> t11 = 87

Therefore, eleventh term in the sequence is 87.

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The probability that your call to a service line is answered in less than 30 seconds is 0.75. Assume that your calls are indepen
vfiekz [6]

Answer:

a) 0.2581

b) 0.4148

c) 17

Step-by-step explanation:

For each call, there are only two possible outcomes. Either they are answered in less than 30 seconds. Or they are not. The probabilities for each call are independent. 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:

p = 0.75

a. If you call 12 times, what is the probability that exactly 9 of your calls are answered within 30 seconds? Round your answer to four decimal places (e.g. 98.7654).

This is P(X = 9) when n = 12. So

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

P(X = 9) = C_{12,9}.(0.75)^{9}.(0.25)^{3} = 0.2581

b. If you call 20 times, what is the probability that at least 16 calls are answered in less than 30 seconds? Round your answer to four decimal places (e.g. 98.7654).

This is P(X \geq 16) when n = 20

So

P(X \geq 16) = P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)

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

P(X = 16) = C_{20,16}.(0.75)^{16}.(0.25)^{4} = 0.1897

P(X = 17) = C_{20,17}.(0.75)^{17}.(0.25)^{3} = 0.1339

P(X = 18) = C_{20,18}.(0.75)^{18}.(0.25)^{2} = 0.0669

P(X = 19) = C_{20,19}.(0.75)^{19}.(0.25)^{1} = 0.0211

P(X = 20) = C_{20,20}.(0.75)^{20}.(0.25)^{0} = 0.0032

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P(X \geq 16) = P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) = 0.1897 + 0.1339 + 0.0669 + 0.0211 + 0.0032 = 0.4148

c. If you call 22 times, what is the mean number of calls that are answered in less than 30 seconds? Round your answer to the nearest integer.

The expected value of the binomial distribution is:

E(X) = np

In this question, we have n = 22

So

E(X) = 22*0.75 = 16.5

The closest integer to 16.5 is 17.

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How many micrograms are equal to 0.00052 grams
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