Which of the following represents the total cost x shirts bought at a cost of (x+5) dollars each?

nirvana33 [79]

Answer:

m∠CDE < m∠DEC < m∠ECD

OR

m∠D < m∠E < m∠C

Step-by-step explanation:

According to the angle-side inequalities theorem: In a triangle, the angle opposite to a longer side is larger, and vise versa, the side opposite to a larger angle is longer.

So, the order from smallest to largest is:

m∠CDE < m∠DEC < m∠ECD

OR

m∠D < m∠E < m∠C

5 0

3 years ago

Which set represents the same relation as the graph below?

Svetllana [295]

Answer:

start with root and minus with 3 and 8 and answer show

5 0

2 years ago

Which box plot correctly displays the data set with a maximum of 48, a minimum of 27, a median of 36, an upper quartile of 45, a

katovenus [111]

Answer:

A.

Step-by-step explanation:

A box-and-whisker plot. Hope this helps! :)

7 0

3 years ago

Read 2 more answers

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

PLEASE HELP

Kitty [74]

Answer:

You don't need to create 3 posts for a single question. I already answered it in your most recent one.

8 0

1 year ago