(b^2)^o=b^8 <br> a 6<br> b 4<br> c 3<br> d 1/16

A square is 7 cm on each side what is the area

Zina [86]

Answer:

49 cm²

Step-by-step explanation:

Formula for area:

a = l (length) x w (width)

if its 7 cm on each side, then the length and width will be 7

7*7 = 49 cm²

5 0

3 years ago

At the fair 4 employees were paid $8.75 an hour to work at a funnel cake stand. They worked 8 hours at regular pay then for 5 ho

nordsb [41]

Answer:

$125.25

Step-by-step explanation:

8.75*h+11.25*e

h is hours regular pay and e is extra pay for overtime. Since its an extra $2.50 for overtime, add that to the regular pay. Put in the hours:

$8.75*8+11.25*5$

Simplify by multiplying 8.75 by 8

$70+11.25*5$

Simplify by multiplying 11.25 by 5

$70+56.25$

Add

$70+56.25=126.25$

Since it says how much did each earn, as in individually, leave it as it is (unless regular pay was $2.18 an hour, which would be a rip-off). I'm pretty sure including the number of employees was meant to throw you off.

7 0

3 years ago

Which is the simplified answer

IceJOKER [234]

Answer:

C. Can you plz mark me brainliest?

Step-by-step explanation:

5 0

3 years ago

Please help ASAP 25 pts + brainliest to right/best answer

Whitepunk [10]

The answer seems to be D!!

4 0

3 years ago

Read 2 more answers

The arrivals of clients at a service firm in Santa Clara is a random variable from Poisson distribution with rate 2 arrivals per

ICE Princess25 [194]

Answer:

1.76% probability that in one hour more than 5 clients arrive

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

The arrivals of clients at a service firm in Santa Clara is a random variable from Poisson distribution with rate 2 arrivals per hour.

This means that $\mu = 2$

What is the probability that in one hour more than 5 clients arrive

Either 5 or less clients arrive, or more than 5 do. The sum of the probabilities of these events is decimal 1. So

$P(X \leq 5) + P(X > 5) = 1$

We want P(X > 5). So

$P(X > 5) = 1 - P(X \leq 5)$

In which

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-2}*2^{0}}{(0)!} = 0.1353$

$P(X = 1) = \frac{e^{-2}*2^{1}}{(1)!} = 0.2707$

$P(X = 2) = \frac{e^{-2}*2^{2}}{(2)!} = 0.2707$

$P(X = 3) = \frac{e^{-2}*2^{3}}{(3)!} = 0.1804$

$P(X = 4) = \frac{e^{-2}*2^{4}}{(4)!} = 0.0902$

$P(X = 5) = \frac{e^{-2}*2^{5}}{(5)!} = 0.0361$

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.1353 + 0.2702 + 0.2702 + 0.1804 + 0.0902 + 0.0361 = 0.9824$

$P(X > 5) = 1 - P(X \leq 5) = 1 - 0.9824 = 0.0176$

1.76% probability that in one hour more than 5 clients arrive

8 0

3 years ago

Read 2 more answers

(b^2)^o=b^8 a 6 b 4 c 3 d 1/16