2 years ago

What is the measure of RCD in the figure below?

Mathematics

1 answer:

Juli2301 [7.4K]2 years ago

4 0

I think the answer to this problem is A

You might be interested in

Kim ate 3/10 of a bag of chips, sandy ate 3/7 of a bag, and jeff ate 3/9 of a bag. who ate the most

aksik [14]

Answer:

sandy

Step-by-step explanation:

3/10 = .3

3/7=.43

3/9=.333333

sandy then jeff then kim

6 0

3 years ago

Read 2 more answers

The average (arithmetic mean) weight of the students in the French Club is 150 pounds, and the average weight of the students in

soldier1979 [14.2K]

20 members are in the spanish club

6 0

3 years ago

If a = -9 and b = 4, what is the value of a + b ?<br> A. -5<br> B. 5 <br> C. -13 <br> D. 13

madreJ [45]

Answer:

A. -5

Step-by-step explanation:

-9 + 4 = -5

hope this helps

7 0

3 years ago

QUICK PLEASE

MA_775_DIABLO [31]

I think it might be D

The size of the sample space is the total number of possible outcomes. For example, when you roll 1 die, the sample space is 1, 2, 3, 4, 5, or 6. So the size of the sample space is 6. Then you need to determine the size of the event space.

7 0

3 years ago

Read 2 more answers

Simple random sampling uses a sample of size n from a population of size N to obtain data that can be used to make inferences ab

OLEGan [10]

Answer: 6,77,040 different random samples of four accounts are possible.

Step-by-step explanation:

We have to calculate the different random samples of four accounts are possible from the population of 65 bank accounts.

we are using combination formula for calculating various combination of four accounts:

= $^{n}C_{r}$

Where,

n - total population of bank accounts

r - random samples drawn from the population

= $^{65}C_{4}$

= $\frac{n!}{r!(n-r)!}$

= $\frac{65!}{4!(65-4)!}$

= $\frac{65!}{4!(61)!}$

= 6,77,040 different random samples of four accounts are possible.

5 0

3 years ago