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2 years ago

Find the GCF of 60, 96, and 156.

Mathematics

1 answer:

Temka [501]2 years ago

8 0

Answer:

GCF = 12

for the values 60, 96, 156

Step-by-step explanation:

Solution by Factorization:

The factors of 60 are: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60

The factors of 96 are: 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96

The factors of 156 are: 1, 2, 3, 4, 6, 12, 13, 26, 39, 52, 78, 156

Then the greatest common factor is 12.

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A skydiver weighing 180lbs (including equipment) falls vertically downward from an altitude of 5000 ft and opens the parachute a

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m=180[lb]=81.6[kg]

g=9.81[m/s2]

b=0.75[kg⋅m/s2⋅s/ft]=0.2286[kg/s]

The solution that the friction provides is v(t)=3501.7[m/s]⋅(1−e−0.00280[1/s]⋅t), where I get 96.69[m/s]=317.2[ft/s]. I am hoping that this answer has satisfied your query about this specific question.

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3 years ago

the temperature in Bloomington on Sunday was 32.6°C. On Monday, the temperature changed by -8.25°C. What was the temperature on

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Step-by-step explanation:

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3 years ago

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3 years ago

A section of a tessellated plane is shown. Which type of symmetry does the tessellated plane have?

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3 years ago

Read 2 more answers

The probability that an Oxnard University student is carrying a backpack is .70. If 10 students are observed at random, what is

Rus_ich [418]

Answer:

35.03% probability that fewer than 7 will be carrying backpacks

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they are carrying a backpack, or they are not. The probability of a student carrying a backpack is independent from other students. 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.

The probability that an Oxnard University student is carrying a backpack is .70.

This means that $p = 0.7$

If 10 students are observed at random, what is the probability that fewer than 7 will be carrying backpacks

This is $P(X < 7)$ when $n = 10$ . So

$P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

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

$P(X = 0) = C_{10,0}.(0.7)^{0}.(0.3)^{10} = 0.000006$

$P(X = 1) = C_{10,1}.(0.7)^{1}.(0.3)^{9} = 0.0001$

$P(X = 2) = C_{10,2}.(0.7)^{2}.(0.3)^{8} = 0.0014$

$P(X = 3) = C_{10,3}.(0.7)^{3}.(0.3)^{7} = 0.0090$

$P(X = 4) = C_{10,4}.(0.7)^{4}.(0.3)^{6} = 0.0368$

$P(X = 5) = C_{10,5}.(0.7)^{5}.(0.3)^{5} = 0.1029$

$P(X = 6) = C_{10,6}.(0.7)^{6}.(0.3)^{4} = 0.2001$

$P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.000006 + 0.0001 + 0.0014 + 0.0090 + 0.0368 + 0.1029 + 0.2001 = 0.3503$

35.03% probability that fewer than 7 will be carrying backpacks

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4 years ago