The vertices of a right triangle are (-3, 5), (4, -1), and (4, y). What is the value of y?

Find the volume of a cylinder with height 12 cm and diameter 7 cm.

Svet_ta [14]

3 0

3 years ago

Harvey bought a. Frame in which he put his family picture in what is the area of the frame not covered by the picture

9966 [12]

Answer:

Area of the frame not covered with photo is equal to 129 square inches

Step-by-step explanation:

Given: Image of a frame

To find: Area of the frame not covered by the picture

Solution:

Area of rectangle = length of the rectangle × breadth of rectangle

Area of the frame = length of frame × breadth of frame = 23 × 15 = 345 square inches

Area of the frame covered with photo = 18 × 12 = 216 square inches

Area of the frame not covered with photo = Area of the frame - Area of the frame covered with photo = 345 - 216 = 129 square inches

8 0

3 years ago

Leah was supposed to attend a party at 3:15 pm. She ran into car trouble that delayed her 35 minutes. Once she got her car fixed

Anit [1.1K]

Answer:

4:10

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

What is polynomail area

erastova [34]

Answer:

2x²-8 if it doesnt give you x= if it gives you an x= then comment it and ill work it out.

Step-by-step explanation:

it was pretty simple

5 0

2 years ago

In a class of 19 students, 3 are math majors. A group of four students is chosen at random. (Round your answers to four decimal

KatRina [158]

Answer:

(a) The probability is 0.4696

(b) The probability is 0.5304

(c) The probability is 0.0929

Step-by-step explanation:

The total number of ways in which we can select k elements from a group n elements is calculate as:

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

So, the number of ways in which we can select four students from a group of 19 students is:

$19C4=\frac{19!}{4!(19-4)!}=3,876$

On the other hand, the number of ways in which we can select four students with no math majors is:

$(16C4)*(3C0)=(\frac{16!}{4!(16-4)!})*(\frac{3!}{0!(3-0)!})=1820$

Because, we are going to select 4 students form the 16 students that aren't math majors and select 0 students from the 3 students that are majors.

At the same way, the number of ways in which we can select four students with one, two and three math majors are 1680, 360 and 16 respectively, and they are calculated as:

$(16C3)*(3C1)=(\frac{16!}{3!(16-3)!})*(\frac{3!}{1!(3-1)!})=1680$

$(16C2)*(3C2)=(\frac{16!}{2!(16-2)!})*(\frac{3!}{2!(3-1)!})=360$

$(16C1)*(3C3)=(\frac{16!}{1!(16-1)!})*(\frac{3!}{3!(3-3)!})=16$

Then, the probability that the group has no math majors is:

$P=\frac{1820}{3876} =0.4696$

The probability that the group has at least one math major is:

$P=\frac{1680+360+16}{3876} =0.5304$

The probability that the group has exactly two math majors is:

$P=\frac{360}{3876} =0.0929$

6 0

3 years ago