SOMEONE PLZ HELP ME

Mathematics

1 answer:

Bingel [31]3 years ago

5 0

If the perimeter is 16, the side length is 4 units.

To find the area of the shaded sections, you will use the fractional part that each section represents and multiply that fraction by the total area of the square.

Area of DEA is 1/4 of 16 square Units (4 x 4).

1/4 x 16 = 4 square units

Area of EFB is 1/8 of 16 square units.

1/8 x 16 = 2 square units

4 square units +2 square units equals 6 square units.

The area of the shaded region is 6 square units.

You might be interested in

Find the volume of the pyramid. Round your answer to the nearest tenth.

algol13

Answer:

from my opinion second option is right

8 0

3 years ago

How would I solve for V?<br><img src="https://tex.z-dn.net/?f=ke%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20mv%5E%7B2%7D%20" id="TexForm

Shkiper50 [21]

(i just changed my answer)

sqrt is the square root of

v=sqrt(2(ke)/m)

4 0

3 years ago

Example 2: Combining Use of the Multiplication and Addition Rules

bulgar [2K]

Answer:

The probability of getting 6's on both cubes is $\frac{1}{36}$ .

The probability that the total score is at least 11 is $\frac{1}{12}$ .

Step-by-step explanation:

Consider the provided information.

A red cube has faces labeled 1 through 6, and a blue cube has faces labeled in the same way.

Part (A) Both cubes show 6’s.

Probability of getting 6 on red cube is $\frac{1}{6}$

Probability of getting 6 on blue cube is $\frac{1}{6}$

Thus, the probability of getting 6's on both cubes is:

$P(\text{Both 6's})=\frac{1}{6}\times\frac{1}{6}=\frac{1}{36}$

Hence, the probability of getting 6's on both cubes is $\frac{1}{36}$ .

Part (B) The total score is at least 11.

The possible number of outcomes in which total score is at least 11 is:

Red shows 6 and Blue shows 5.

Blue shows 6 and Red shows 5.

Blue shows 6 and Red shows 6.

Thus, the probability of total score is at least 11.

$P(\text{Total is at least 11})=\frac{1}{6}\times\frac{1}{6}+\frac{1}{6}\times\frac{1}{6}+\frac{1}{6}\times\frac{1}{6}\\P(\text{Total is at least 11})=\frac{1}{12}$