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qwelly [4]
3 years ago
13

What is 7.25 sales tax on 275.00

Mathematics
1 answer:
german3 years ago
6 0

Answer:

$19.94

Step-by-step explanation:

275 * 0.0725 = 19.9375, round it to 19.94

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How many positive odd factors of 48 are greater than 2 and less than 10
Alexeev081 [22]

Answer:

Just 1 positive, odd factor

Step-by-step explanation:

48 has many factors. But not many odd factors. Only 3 is odd and bigger than 2 and less than 10.

So, just one odd, positive factor.

7 0
2 years ago
Dylan plans to eat 1/5 of his candy bar. his 4 friends want him to share the rest equally. show how Dylan and his friends can ea
Paul [167]
They would each get 1/5 of the candy bar
5 0
3 years ago
Which expression shows a way to find 30% of 10
sdas [7]

Answer:

3

Step-by-step explanation:

30% of 10 would be 30 divided by 10

8 0
3 years ago
Larry earns $11.25/hour selling televisions. He also earns 4.5% commission on his sales. What is his
monitta

Answer:

C) $573.75

Step-by-step explanation:

We can come up with the equation of this Gross Weekly pay as:

P(t)=11.25t+0.045S

Where P is the amount Larry will get paid, t is the time worked and S is the amount of sales he generated that week.  This was derived from the fact that 11.25 is the slope and 0.045S is the y-intercept.  By plugging in those values we have:

P(35)=11.25(35)+0.045(4000)=573.75

C) $573.75

4 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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