Ask question

Ask question

All categories

Anestetic [448]

3 years ago

15

Harry earns an hourly rate of $15. For each additional hour over 40 hours, he earns 150% percent of his regular hourly rate. Las

t week Harry earned$847.50 working 40 regular hour and some overtime hours. How many total hours did Harry work last week.

1 answer:

Yuri [45]3 years ago

7 0

Answer:

51 hours

$15 x 40 = $600

$15 x 1.50 = $22.5

$22.5 x 11 = $247.5

$600 + $247.5 = $847.5

40 + 11 = 51 hours

You might be interested in

What is 10(n−2p+2) because i dont know the answer

Zigmanuir [339]

10n - 20p + 20

could i get a heart and branliest

6 0

2 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

2 years ago

Read 2 more answers

In 1981, the Australian humpback whale population was 350 and has increased at a rate of 14% each year since then.)

dimulka [17.4K]

Answer:

In 1981, the Australian humpback whale population was 350

Po = Initial population = 350

rate of increase = 14% annually

P(t) = Po*(1.14)^t

P(t) = 350*(1.14)^t

Where

t = number of years that have passed since 1981

Year 2000

2000 - 1981 = 19 years

P(19) = 350*(1.14)^19

P(19) = 350*12.055

P(19) = 4219.49

P(19) ≈ 4219

Year 2018

2018 - 1981 = 37 years

P(37) = 350*(1.14)^37

P(37) = 350*127.4909

P(37) = 44621.84

P(37) ≈ 44622

There would be about 44622 humpback whales in the year 2018

5 0

3 years ago

Find the distance (-4,5) and (7,18)

seropon [69]

Answer:

$\sqrt{290}\ units$

Step-by-step explanation:

The distance between points $(x_1,y_1)$ and $(x_2,y_2)$ is

$\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

Hence, the distance between points $(-4,5)$ and $(7,18)$ is

$\sqrt{(-4-7)^2+(5-18)^2}\\ \\=\sqrt{(-11)^2+(-13)^2}\\ \\=\sqrt{121+169}\\ \\=\sqrt{290}\ units$

3 0

3 years ago

What is 7 divided by 18

Aleks04 [339]

7 ÷ 18 is 0.38888888888

:)

6 0

3 years ago

Read 2 more answers