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patriot [66]
3 years ago
8

Let f(x)=−2x−7. Write a function g whose graph is a translation 6 units to the right of the graph of f

Mathematics
1 answer:
Andreyy893 years ago
5 0
There must be More details for the problem
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A recently negotiated union contract allows workers in a shipping department 24 minutes for rest, 10 minutes for personal time,
Alik [6]

Answer:7.125 min

Step-by-step explanation:

Given

Resting time =24 min

Personal time=10 min

Delay=14 min

Observed Time=6min/cycle

Performance Rating=95%

Allowance Time=Rest time+Personal time+delay

Allowance time=24+10+14=48 min

Allowance %=\frac{Allowance Time}{Total Time}=\frac{48}{4\times 60}=20 \%

Allowance Factor=\frac{1}{1-0.20}=\frac{5}{4}

Calculate the Normal Time

Normal time=Observed time\timespersonal Time=6\times 0.95=5.7 min

Standard time=Normal time \times Allowance Factor

Standard time =NT\times Allowance factor=5.7\times 1.25=7.125 min

3 0
3 years ago
A line passes through the point (–2, 7) and has a slope of –5.
kakasveta [241]

Answer:

it is -7 because the 2 make it like that

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Divide. 83 ÷ 9 Enter your answer by filling in the boxes. __R __
forsale [732]

9R2

83/9 = 9.2222222

9*9 = 81

83-81 = 2

5 0
2 years ago
Read 2 more answers
Jaxon has $65 in his savings account. He deposits $15 every week. His father also deposits $25 into the account every time Jaxon
Grace [21]
Part A:
A coefficient can be either '15' or '25'.
A variable can be either 'w' or 'm'.
A constant is 65.

Part B:
Simply substitute, or plug-in, the numbers and solve.
***Step 1:
65+15w+25m --> 65+15(20)+25(3)
You do this because you are substituting the 'w' for the number
of weeks that Jaxon saved up for, which is 20, and the 'm' for the number
of times that Jaxon mowed the lawn, which is 3.
***Step 2:
65+15(20)+25(3) --> 65+300+75
Begin to solve, using PEMDAS, or whichever acronym you learned.
Remember, if you are using PEMDAS, recall that the order is Parenthesis,
Exponents, Multiplication/Division (whichever comes first), and
Addition/Subtraction (whichever comes first). Here, I checked for parenthesis.
I did find parenthesis, however, they do not have any expressions inside of
them, meaning that these parenthesis are for multiplying, and not for stating
order. So, you skip parenthesis. Next, you check for exponents, which you
find none of, so you skip over that. Now, we get to multiplying/dividing, so
you multiply the 15 and the 20 to get 300, and the 25 and 3 to get 75.
***Step 3:
65+300+75 --> 440
Now, we get to addition. You simply add everything up to get your final
answer: $440.

Part C:
If Jaxon had $75, then yes, the coefficients would change.
By subtracting $65 from $75, we can see that the total amount of money
from Jaxon's deposits and his lawn-mowing money is $10. Jaxon already
deposits $15 a week, meaning that, while using the current equation, Jaxon
CANNOT have $75 in his bank account. We can change the equation
so that Jaxon is able to have $75 in his savings account. You can change
the coefficient of 15 to 10, and the other coefficient of 25 to 0.
Now Jaxon is able to have $75 in his savings account.
8 0
3 years ago
In a recent survey of college professors, it was found that the average amount of money spent on entertainment each week was nor
Alenkinab [10]

Answer:

0.0918

Step-by-step explanation:

We know that the average amount of money spent on entertainment is normally distributed with mean=μ=95.25 and standard deviation=σ=27.32.

The mean and standard deviation of average spending of sample size 25 are

μxbar=μ=95.25

σxbar=σ/√n=27.32/√25=27.32/5=5.464.

So, the average spending of a sample of 25 randomly-selected professors is normally distributed with mean=μ=95.25 and standard deviation=σ=27.32.

The z-score associated with average spending $102.5

Z=[Xbar-μxbar]/σxbar

Z=[102.5-95.25]/5.464

Z=7.25/5.464

Z=1.3269=1.33

We have to find P(Xbar>102.5).

P(Xbar>102.5)=P(Z>1.33)

P(Xbar>102.5)=P(0<Z<∞)-P(0<Z<1.33)

P(Xbar>102.5)=0.5-0.4082

P(Xbar>102.5)=0.0918.

Thus, the probability that the average spending of a sample of 25 randomly-selected professors will exceed $102.5 is 0.0918.

5 0
3 years ago
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