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3 years ago

Find the smallest value of m such that the LCM of m and 34 is 374.

Mathematics

1 answer:

wlad13 [49]3 years ago

8 0

Answer:

<em>11</em>

first you use a number tree

34 374

/\ /\

2 17 2 187

11 17

express the numbers in prime factors

374=2×11×17

34 =2×17

in finding LCM you use the highest prime factors in this case the highest prime factors are 2, 11 and 17

So m factors are 2 and 11

which means m is 2×11=22

22 in its smallest form is 11

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3 years ago

Josh attempted 56 free throws during the basketball season. He made 57 of them. How many free throws did he make?

Scorpion4ik [409]

Answer:

He made approximately 32 free throws.

Step-by-step explanation:

Assuming that there's a typo in the question and that he made 57% of his attempted free throws, then we can solve it as shown below:

We can apply a rule of three in order to calculate the number of free throws he made. This is done as follows:

56 free throws -> 100%

x free throws -> 57%

56/x = 100/57

100*x = 57*56

x = 57*56/100 = 31.92

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3 years ago

In a MBS first year class, there are three sections each including 20 students. In the first section, there are 10 boys and 10 g

KIM [24]

Answer:

$3.52 \times 10^{-9} = 3.52 \times 10^{-7}\%$ probability that all the 15 students selected are girls

Step-by-step explanation:

The selection is from a sample without replacement, so we use the hypergeometric distribution to solve this question.

Hypergeometric distribution:

The probability of x sucesses is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

All girls from the first group:

20 students, so $N = 20$

10 girls, so $k = 10$

5 students will be selected, so $n = 5$

We want all of them to be girls, so we find P(X = 5).

$P_1 = P(X = 5) = h(5,20,5,10) = \frac{C_{10,5}*C_{10,5}}{C_{20,5}} = 0.0163$

All girls from the second group:

20 students, so $N = 20$

5 girls, so $k = 5$

5 students will be selected, so $n = 5$

We want all of them to be girls, so we find P(X = 5).

$P_2 = P(X = 5) = h(5,20,5,5) = \frac{C_{5,5}*C_{15,5}}{C_{20,5}} = 0.00006$

All girls from the third group:

20 students, so $N = 20$

8 girls, so $k = 8$

5 students will be selected, so $n = 5$

We want all of them to be girls, so we find P(X = 5).

$P_3 = P(X = 5) = h(5,20,5,8) = \frac{C_{8,5}*C_{12,5}}{C_{20,5}} = 0.0036$

All 15 students are girls:

Groups are independent, so we multiply the probabilities:

$P = P_1*P_2*P_3 = 0.0163*0.00006*0.0036 = 3.52 \times 10^{-9}$

$3.52 \times 10^{-9} = 3.52 \times 10^{-7}\%$ probability that all the 15 students selected are girls

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3 years ago