How do you work out 3^2+2^3 indices

There are 16 friends sharing for cupcakes if they invited cupcakes even they said that each get four pieces of a cupcake how man

emmainna [20.7K]

You have to cut them into forth sized pieces. Each of 16 friends will get 1/4 of a cupcake.

8 0

3 years ago

Which of the Following is an example of commutative property of addition.

katen-ka-za [31]

The answer is
A. 3+9=9+3

8 0

3 years ago

In a bakery, a bag of bread rolls cost x dollars. A loaf of multi-grain bread costs $2 more than a bag of bread rolls. Mr. Lopez

Vaselesa [24]

Step-by-step explanation:

Cost of bag of bread rolls is x

Multigrain is (+2) more.

So (x+2)

Y can be the number of multigrain loves Mr. Lopez buys.

Answer: y=50/(x+2)

6 0

2 years ago

Haroldo, Xerxes, Regina, Shaindel, Murray, Norah, Stav, Zeke, Cam, and Georgia are invited to a dinner party. They arrive in a r

alexira [117]

Answer: The probability is P = 0.056

Step-by-step explanation:

The invited ones are:

Haroldo, Xerxes, Regina, Shaindel, Murray, Norah, Stav, Zeke, Cam, and Georgia

So we have a total of 10 persons.

The case where Xerxes arrives first and Regina arrives last is:

We have 10 slots, each slot corresponds to who arrived in which time.

For the first slot we have only one option, this is Xeres.

For the next slot we have 8 options (because Xeres already arrived, and Regina must arrive at last)

For the next slot we have 7 options.

for the next one we have 6 options, and so on.

So the combinations are:

C = 1*8*7*6*5*4*3*2*1*1 = 40320

Now, the total number of combinations is:

for the first slot we have 10 options, for the second we have 9 options and so on.

c = 10*9*8*7*6*5*4*3*2*1 = 725760

The probability is the combinations where Xerxes arrives first and Regina rrives last divided the total number of probabilities.

P = C/c = 0.056

7 0

3 years ago

There are 3 red and 8 green balls in a bag. If you randomly choose balls one at a time, with replacement, what is the probabilit

Kaylis [27]

So, the total number of balls is 11. We want to pick 2 red balls and 1 green ball. WLOG (since order doesnt matter here), we can say he picks red, green, red. That means on his first pick, he has a $\frac{3}{11}$ chance of picking the red ball, and he places it back in the bag. The probability of picking a green ball is $\frac{8}{11}$ , and then he places the ball back in the bag. The probability of picking the last red ball is the same as the last red ball example, and we simply multiply the probabilities together as per the multiplication rule to get:

$\frac{3}{11}\frac{3}{11}\frac{8}{11}=\frac{3*3*8}{11^3}=\frac{72}{1331}$

Now, without replacement the order does matter. He picks a red ball, a red ball then a green ball. The probability of picking the first red ball is $\frac{2}{11}$ , and the probability of picking the second red ball is $\frac{1}{10}$ and the probability of picking the green ball is $\frac{1}{9}$ . We want to multiply thm again, as per the multiplication rule like the last problem.

$\frac{2}{11}*\frac{1}{10}*\frac{1}{9}=\frac{1}{11*5*9}=\frac{1}{495}$

5 0

3 years ago