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alexdok [17]
3 years ago
7

Jan Quint earns $11.00 an hour at her job and is entitled to time-and-a-half for overtime, and double time on holidays. Last wee

k she worked 40 hours of regular time, 9 hours of overtime, and 12 hours of holiday time. How much did she earn last week?
A. $720.50
B. $612.00
C. $852.50
D. $803.00
Please explain the answer..
Mathematics
2 answers:
tamaranim1 [39]3 years ago
8 0
Well, if it's $11 per hour, and you need to find out how much she got paid in total, you would add all of da hours together, which would be 61, and multiply 11 times 62 = $671, so I would round dat to D., $803.00.
I hope I helped! =D
marysya [2.9K]3 years ago
6 0
Time and a half mean
example, if he work for 1 hour normal, then he get 1 hour pay
but time and a half, is if he works 1 hour overitme, he gets paid for 1.5 time
double time is work 1 hour, get paid 2 hours

basically, multiply all normal hours by 1, multpliy al overtime hours by 1.5 and multiply al holiday hours by 2

40 hours normal
9 hours overtime
12 hours holidy
40*1=40
9*1.5=13.5
12*3=24
add up
40+13.5+24=77.5 hours counted

now 11 per our
11 times 77.5=852.50

C is answer

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Answer:

a) Our random variable X="number of tests taken until the individual passes" follows a geomteric distribution with probability of success p=0.25

For this case the probability mass function would be given by:

P(X= k) = (1-p)^{k-1} p , k = 1,2,3,...

b) P(X \leq 3) = P(X=1) +P(X=2) +P(X=3)

P(X= 1) = (1-0.25)^{1-1} *0.25 = 0.25

P(X= 2) = (1-0.25)^{2-1} *0.25 = 0.1875

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And adding the values we got:

P(X \leq 3) =0.25+0.1875+0.1406=0.578

c) P(X \geq 5) = 1-P(X

And we can find the individual probabilities:

P(X= 1) = (1-0.25)^{1-1} *0.25 = 0.25

P(X= 2) = (1-0.25)^{2-1} *0.25 = 0.1875

P(X= 3) = (1-0.25)^{3-1} *0.25 = 0.1406

P(X= 4) = (1-0.25)^{4-1} *0.25 = 0.1055

P(X \geq 5) = 1-[0.25+0.1875+0.1406+0.1055]= 0.316

Step-by-step explanation:

Previous concepts

The geometric distribution represents "the number of failures before you get a success in a series of Bernoulli trials. This discrete probability distribution is represented by the probability density function:"

P(X=x)=(1-p)^{x-1} p

Let X the random variable that measures the number of trials until the first success, we know that X follows this distribution:

X\sim Geo (1-p)

Part a

Our random variable X="number of tests taken until the individual passes" follows a geomteric distribution with probability of success p=0.25

For this case the probability mass function would be given by:

P(X= k) = (1-p)^{k-1} p , k = 1,2,3,...

Part b

We want this probability:

P(X \leq 3) = P(X=1) +P(X=2) +P(X=3)

We find the individual probabilities like this:

P(X= 1) = (1-0.25)^{1-1} *0.25 = 0.25

P(X= 2) = (1-0.25)^{2-1} *0.25 = 0.1875

P(X= 3) = (1-0.25)^{3-1} *0.25 = 0.1406

And adding the values we got:

P(X \leq 3) =0.25+0.1875+0.1406=0.578

Part c

For this case we want this probability:

P(X \geq 5)

And we can use the complement rule like this:

P(X \geq 5) = 1-P(X

And we can find the individual probabilities:

P(X= 1) = (1-0.25)^{1-1} *0.25 = 0.25

P(X= 2) = (1-0.25)^{2-1} *0.25 = 0.1875

P(X= 3) = (1-0.25)^{3-1} *0.25 = 0.1406

P(X= 4) = (1-0.25)^{4-1} *0.25 = 0.1055

P(X \geq 5) = 1-[0.25+0.1875+0.1406+0.1055]= 0.316

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3 years ago
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