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3 years ago

(08.04 MC)

Mathematics

1 answer:

skelet666 [1.2K]3 years ago

7 0

Answer:

292,968

Step-by-step explanation:

As we know,

Sum of a geometric sequence (S) = $\frac{a(1-r^{n}) }{(1-r)}$

where,

a = first term of sequence,

r = the constant ratio,

n = number of terms in sequence.

So, according to the question,

a = 3,

r = 5,

n = 8.

by substituting the values in the above formula, we get;

⇒ $S=\frac{3(1-5^{8}) }{(1-5)}$

⇒ $292,968$

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<h3>Answer: 4900 (choice D)</h3>

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Work Shown:

1 km = 1000 m

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Answer:

(A) The percent of the problems takes more than 5 minutes to resolve is 60.87%.

(B) The problem solving time will be 50% as long as the difference between the two end points is 5.75.

(D) Mean = 6.25 minutes, Standard deviation = 3.32 minutes

Step-by-step explanation:

Let the random variable X represent the time it takes the technician to resolve the problem.

The random variable X follows an Uniform distribution with parameters a = 0.50 minutes and b = 12 minutes.

The probability density function of X is:

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(A)

Compute the probability that the problems takes more than 5 minutes to resolve as follows:

$P(X>5)=\int\limits^{12}_{5}{\frac{1}{12-0.50}}\, dx$

$=\frac{1}{11.50}\cdot \int\limits^{12}_{5}{1}\, dx \\\\=\frac{1}{11.50}\cdot [x]\limits^{12}_{5}\\\\=\frac{1}{11.50}\cdot [12-5]\\\\=\frac{7}{11.50}\\\\=0.608696\\\\\approx 0.6087$

Thus, the percent of the problems takes more than 5 minutes to resolve is 60.87%.

(B)

Let the middle 50% of the problem-solving times be between u and v.

Then,

P (u < X < v) = 0.50

$\int\limits^{v}_{u}{\frac{1}{12-0.50}}\, dx=0.50\\\\\frac{1}{11.50}\cdot \int\limits^{v}_{u}{1}\, dx=0.50\\\\\frac{v-u}{11.50}=0.50\\\\(v-u)=5.75$

Thus, the problem solving time will be 50% as long as the difference between the two end points is 5.75.

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The interval in which the technician can solve the problem is 30 seconds to 12 minutes.

So, the values of a and b in minutes is:

a = 30 seconds = 0.50 minutes

b = 12 minutes.

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The mean time is:

$\mu=\frac{a+b}{2}=\frac{0.50+12}{2}=6.25\ \text{minutes}$

The standard deviation of the time is:

$\sigma=\sqrt{\frac{(b-a)^{2}}{12}}=\sqrt{\frac{(12-0.50)^{2}}{12}}=3.3197\approx 3.32\ \text{minutes}$

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