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Xelga [282]
4 years ago
15

One half of negative five eighths

Mathematics
2 answers:
likoan [24]4 years ago
7 0
First make negative five eighths a decimal by dividing the numerator by the denominator (-.625).
Now divide that decimal in half (-.3125)
Last take that decimal and put it back into fraction form (-5/16).
One half of negative five eighths is -5/16.
evablogger [386]4 years ago
6 0
-0.3125 is the answer, but you probably have to round
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Please help!! Will give brainliest! Giving lots of points!
olga nikolaevna [1]

Answer:

d) (9*3/4)+y

Step-by-step explanation:

b) 9+y*(3/4)

This expression is not equivalent, as the original expression, y+9*\frac{3}{4} , has 9 multiplied to \frac{3}{4} instead of y. You can interchange values like this when it comes to addition, but not multiplication.

c) (y+9)(3/4)

This expression is also not equivalent. After opening up the parentheses, it would again result in y being multiplied to \frac{3}{4}.

(y+9)(3/4)\\= y(3/4)+9(3/4)

d) (9*3/4)+y

This expression is equivalent, as 9 and \frac{3}{4} are still being multiplied together and y is added to the product. We can actually rearrange this to look like the original expression:

(9*3/4)+y\\=y+(9*3/4)\\=y+(9*\frac{3}{4} )\\=y+9*\frac{3}{4}

I hope this helps!

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Step-by-step explanation:

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Choose the best definition for the following term: exponent
stealth61 [152]
The answer is B - The number or value being raised to a power

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Step-by-step explanation:

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2 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
4 years ago
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