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Tom [10]
2 years ago
10

B) Select all expressions that can be used to find 85% of 280.

Mathematics
1 answer:
abruzzese [7]2 years ago
4 0

Answer:

238

85%of 280 is 238

Step-by-step explanation:

hope this helps

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The shop declare a discount of 20% and later a discount of 15% after
Zolol [24]
Let, the original amount = x
First discount = x * 0.20 = 0.20x

Amount after first discount = x - 0.20x = 0.80x
Second discount = 0.80x * 0.15 = 0.12x

Amount after second discount = 0.80x - 0.12x = 0.68x
Difference would be: x-0.68x = 0.32x

So, Your Answer would be 32%

Hope this helps!
8 0
3 years ago
Joshua wrote 13 articles for the school newspaper this year. Paulette wrote 7 more articles than Joshua. Jeff wrote as many arti
Keith_Richards [23]
53 articles? Because josh wrote 13 plus 20 that Paulette wrote then Jeff wrote as many as Paulette so another 20.
6 0
3 years ago
What is the ratio of 364:78
Klio2033 [76]
You'd want to simplify that by dividing both terms by the greatest common factor which is the product of the shared primes of both numbers prime factorization...

364=2*2*7*13, 78=2*3*13, so the GCF is 2*13=26, now divide both terms by 26

14:3
8 0
3 years ago
1. Bobby looks in his wallet and sees that he has 2 twenty dollar bills and 1 ten dollar bill. He goes to the store and the chec
Orlov [11]

Answer: a

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
A certain car model has a mean gas mileage of 34 miles per gallon (mpg) with a standard deviation A pizza delivery company buys
zubka84 [21]

Answer:

z =\frac{33.3- 34}{\frac{5}{\sqrt{54}}}= -1.028

z =\frac{34.3- 34}{\frac{5}{\sqrt{54}}}= 0.441

An we can use the normal standard table and the following difference and we got this result:

P(-1.028

Step-by-step explanation:

Assuming this statement to complete the problem "with a standard deviation 5 mpg"

We have the following info given:

\mu = 34 represent the mean

\sigma= 5 represent the deviation

We have a sample size of n = 54 and we want to find this probability:

P(33.3 < \bar X< 34.3)

And for this case since the sample size is large enough >30 we can apply the central limit theorem and then we can use this distribution:

\bar X \sim N(\mu , \frac{\sigma}{\sqrt{n}})

And we can use the z score formula given by:

z=\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And replacing we got:

z =\frac{33.3- 34}{\frac{5}{\sqrt{54}}}= -1.028

z =\frac{34.3- 34}{\frac{5}{\sqrt{54}}}= 0.441

An we can use the normal standard table and the following difference and we got this result:

P(-1.028

7 0
3 years ago
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