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3 years ago

What is the perimeter of the triangle.

Mathematics

1 answer:

Dmitry [639]3 years ago

8 0

36. Find the hypotenuse by squaring the length of the height and length. It gives you 225...so your side C should be 15. Then add all your sides giving you 36

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Mark went to the sandwich shop. He had a choice of 2 types of bread, 3 meats, and 3 drinks. How many different combinations did

kobusy [5.1K]

B(Bread) D(Drink) M(Meat)

B1+D1+M1
B1+D1+M2
B1+D1+M3
B1+D2+M1
B1+D2+M2
B1+D2+M3
B1+D3+M1
B1+D3+M2
B1+D3+M3

B2+D1+M1
B2+D1+M2
B2+D1+M3
B2+D2+M1
B2+D2+M2
B2+D2+M3
B2+D3+M1
B2+D3+M2
B2+D3+M3

Making there be 18 different combinations.

6 0

3 years ago

$0.63 + _______ = $1.00*

Simora [160]

1-.63=.37. so $0.63+$0.37=$1.00

5 0

3 years ago

Read 2 more answers

Factoring a Monomia form of polynomial

AURORKA [14]

Answer:

Factoring a monomial means breaking it down into each individual prime number and/or variable factor. This is also called "expanding" a monomial.

Step-by-step explanation:

4 0

3 years ago

Which inequality is true for x=5

Andreyy89

Answer:

The answer is A) 2x+ 5 > 13

Step-by-step explanation:

4 0

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

7 0

3 years ago