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marusya05 [52]
3 years ago
14

P1. (3+7 points) What is the smallest positive integer with precisely 5 positive divisors? What is the smallest positive integer

with precisely 60 positive divisors? Show your work and reasoning.
Mathematics
1 answer:
qwelly [4]3 years ago
8 0

Answer:

smallest positive integer with 5 positive divisor is 16

smallest positive integer with 60 positive divisor is 5040

Step-by-step explanation:

given data

precisely positive divisors = 5

precisely positive divisors = 60

solution

we take here a^{x} *b^{y} *c^{z} is express as

= (x+1) × (y+1) × (z+1)

so put here now x is 4

and z = y = 0 and a is least integer more than 1 it will be 2

and b and c ≥ 1

and x^{0} is = 1

so a^{x} *b^{y} *c^{z} is

a^{4} is = 2^{4}  = 16  

so smallest positive integer with 5 positive divisor is 16

and

same like 60 positive divisors

dn = ( a1+1 ) × ( a2+1 ) × ( a3+1 )  ............ ( an+1 )

n = p1^{a1} * p2^{a2} * p3^{a3} * ............ pn^{an} *

n = 7 × 5 × 3² × 2^{4}

n = 5040

smallest positive integer with 60 positive divisor is 5040

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3 years ago
Point 6 is on line segment FH. Given FG=8 and GH=11, determine length FH
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Answer:

GH = 20

Step-by-step explanation:

We have that

FH = FG + GH ← substitute values

4x + 8 = x + 5x

4x + 8 = 6x ( subtract 4x from both sides )

8 = 2x ( divide both sides by 2 )

4 = x

Hence GH = 5x = 5 × 4 = 20GH = 20

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2 years ago
A company is manufacturing highway emergency ares. Such ares are sup- posed to burn for an average of 20 minutes. Every hour a s
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Complete question :

A company is manufacturing highway emergency ares. Such ares are sup- posed to burn for an average of 20 minutes. Every hour a sample of ares is collected, and their average burn time is determined. If the manufacturing process is working cor- rectly, there is a 68% chance that the average burn time of the sample will be between 14 minutes and 26 minutes. The quality engineer in charge of the process believes that if 4 of 5 samples fall outside these bounds then this is a signal that the process might not be performing as expected. Each morning the sampling begins anew. Let X denote the number of samples drawn in order to obtain the fourth sample whose average value is outside of the above bounds. Find the probability that for a given morning X = 5, hence there seems to be a problem right away.

Answer:

0.0285

Step-by-step explanation:

We are to find the probability for a given morning that X = 5

From the question, we could see that there is a 68% chance that the average burn time of the sample will be between 14 minutes and 26 minutes and the success of the process is obtained after the fourth or fifth sample.

Thus, r = 4; q = 0.68

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Therefore,

f(x) = \left[\begin{array}{ccc}x-1\\r-1\end{array}\right] (1-p)^x^-^r) (p)^r

Where, x = 5

r = 4

p = 0.32

Substituting figures in the equation, we have:

P(x=5) = \left[\begin{array}{ccc}5-1\\4-1\end{array}\right] (1-0.32)^5^-^4 (0.32)^4

P(x=5) = \left[\begin{array}{ccc}4\\3\end{array}\right] (0.68)(0.01048576)

= 0.0285

The probability that for a given morning X = 5 is 0.0285

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Step-by-step explanation:

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