In triangle ABC, what is the measure of angle C if A = 60°, a = 9, and c = 5?

Which of the following statements must be true about this diagram?<br> Check all that apply.

Alik [6]

Answer:

i would go for a and c

Step-by-step explanation:

7 0

2 years ago

Sum and difference of 5+2.3

professor190 [17]

Answer:

sum: 7.3 difference:2.7

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

64 as a mixed number

Kaylis [27]

I'm perdy sure it's 64/1 but yk it could be something else XD :D good luck!

6 0

2 years ago

Can someone please help me find the side length area, and please draw the shape on how it would look if it were finished

Mumz [18]

The side length is 5 units, and the area is 25 square units

<h3>What are areas?</h3>

The area of a shape is the amount of space it occupies

<h3>What is the area of a square</h3>

The area of a square is the product of its dimensions

From the graph, the coordinates of the square can be assumed to be:

$A = (0,0)$

$B = (4,3)$

The side length of the square is then calculated using the following distance formula

$d = \sqrt{(x_2- x_1)^2 + (y_2 -y_1)^2}$

So, we have:

$AB = \sqrt{(0 - 4)^2 + (0-3)^2}$

$AB = \sqrt{25}$

$AB = 5$

The area of the square is the calculated as:

$Area = AB^2$

This gives

$Area = 5^2$

$Area = 25$

Hence, the side length is 5 units, and the area is 25 square units

See attachment for the diagram of the shape

Read more about areas at:

brainly.com/question/24487155

6 0

2 years ago

A certain company sends 40% of its overnight mail parcels by means of express mail service A1. Of these parcels, 4% arrive after

Harrizon [31]

Answer:

(a) The probability that a randomly selected parcel arrived late is 0.026.

(b) The probability that a parcel was late was being shipped through the overnight mail service A₁ is 0.615.

(c) The probability that a parcel was late was being shipped through the overnight mail service A₂ is 0.192.

(d) The probability that a parcel was late was being shipped through the overnight mail service A₃ is 0.192.

Step-by-step explanation:

Consider the tree diagram below.

(a)

The law of total probability sates that: $P(A)=P(A|B)P(B)+P(A|B')P(B')$

Use the law of total probability to determine the probability of a parcel being late.

$P(L)=P(L|A_{1})P(A_{1})+P(L|A_{2})P(A_{2})+P(L|A_{3})P(A_{3})\\=(0.04\times0.40)+(0.01\times0.50)+(0.05\times0.10)\\=0.026$

Thus, the probability that a randomly selected parcel arrived late is 0.026.

(b)

The conditional probability of an event A provided that another event B has already occurred is:

$P(A|B)=\frac{P(B|A)P(A)}{P(B)}$

Compute the probability that a parcel was late was being shipped through the overnight mail service A₁ as follows:

$P(A_{1}|L)=\frac{P(L|A_{1})P(A_{1})}{P(L)} \\=\frac{0.04\times 0.40}{0.026} \\=0.615$

Thus, the probability that a parcel was late was being shipped through the overnight mail service A₁ is 0.615.

(c)

Compute the probability that a parcel was late was being shipped through the overnight mail service A₂ as follows:

$P(A_{2}|L)=\frac{P(L|A_{2})P(A_{2})}{P(L)} \\=\frac{0.01\times 0.50}{0.026} \\=0.192$

Thus, the probability that a parcel was late was being shipped through the overnight mail service A₂ is 0.192.

(d)

Compute the probability that a parcel was late was being shipped through the overnight mail service A₂ as follows:

$P(A_{3}|L)=\frac{P(L|A_{3})P(A_{3})}{P(L)} \\=\frac{0.05\times 0.10}{0.026} \\=0.192$

Thus, the probability that a parcel was late was being shipped through the overnight mail service A₃ is 0.192.

4 0

2 years ago