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photoshop1234 [79]
3 years ago
6

The eleventh term from the end of the AP 14, 19, 24, …………., 264 is

Mathematics
1 answer:
vazorg [7]3 years ago
6 0

Answer:

<h2>The eleventh term of the sequence is 64</h2>

Step-by-step explanation:

The sequence given is an arithmetic sequence

14, 19, 24, …………., 264

The nth term of an arithmetic sequence is given as;

Tn = a+(n-1)d where;

a is the first term = 14

d is the common difference = 19-14=24-19 = 5

n is the number of terms = 11(since we are to look for the eleventh term of the sequence)

substituting the given values in the formula given;

T11 = 14+(11-1)*5

T11 = 14+10(5)

T11 = 14+50

T11 = 64

The eleventh term of the sequence is 64

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Step-by-step explanation:

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3 years ago
A new security system needs to be evaluated in the airport. The probability of a person being a security hazard is 4%. At the ch
ivolga24 [154]

Answer:

(a) 0.9412

(b) 0.9996 ≈ 1

Step-by-step explanation:

Denote the events a follows:

P = a person passes the security system

H = a person is a security hazard

Given:

P (H) = 0.04,\ P(P^{c}|H^{c})=0.02\ and\ P(P|H)=0.01

Then,

P(H^{c})=1-P(H)=1-0.04=0.96\\P(P|H^{c})=1-P(P|H)=1-0.02=0.98\\

(a)

Compute the probability that a person passes the security system using the total probability rule as follows:

The total probability rule states that: P(A)=P(A|B)P(B)+P(A|B^{c})P(B^{c})

The value of P (P) is:

P(P)=P(P|H)P(H)+P(P|H^{c})P(H^{c})\\=(0.01\times0.04)+(0.98\times0.96)\\=0.9412

Thus, the probability that a person passes the security system is 0.9412.

(b)

Compute the probability that a person who passes through the system is without any security problems as follows:

P(H^{c}|P)=\frac{P(P|H^{c})P(H^{c})}{P(P)} \\=\frac{0.98\times0.96}{0.9412} \\=0.9996\\\approx1

Thus, the probability that a person who passes through the system is without any security problems is approximately 1.

7 0
2 years ago
What is 7(b-2) is b=8
Delvig [45]
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42 ans
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3 years ago
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GalinKa [24]

Answer:

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4 0
3 years ago
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How to factor 1-49c^2
Dima020 [189]
Ok so this is a difference of 2 perfect squares
1=1^2
49c^2=(7c)^2
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a^2-b^2=(a+b)(a-b) so

1^2-(7c)^2=(1+7c)(1-7c)
5 0
2 years ago
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