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tigry1 [53]
3 years ago
14

1. We can use division to solve real life problems, such as:

Mathematics
2 answers:
KonstantinChe [14]3 years ago
8 0
I think the answer is D
iren2701 [21]3 years ago
4 0
Well, let's look at the answer choices.
A) If you want to fond the total cost of two sweaters you would add, not divide. So this isn't it.
B) To find out how much older someone is you would subtract. So this isn't it either.
C) To find out how many desks are needed you would multiply, since you know you need 3 rows of desks with 7 students each (3 x 7). Again, not it.
D) To find out how many pencils each student would get, you'd divide. You would divide 56 by 9. So your answer is D.
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There are about 4200 college presidents in the United States, and they have annual incomes with a distribution that is skewed in
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3 0
3 years ago
???????I don't get it
Marta_Voda [28]
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8 0
3 years ago
Determine what type of model best fits the given situation:
lyudmila [28]

Let value intially be = P

Then it is decreased by 20 %.

So 20% of P = \frac{20}{100} \times P = 0.2P

So after 1 year value is decreased by 0.2P

so value after 1 year will be = P - 0.2P (as its decreased so we will subtract 0.2P from original value P) = 0.8P-------------------------------------(1)

Similarly for 2nd year, this value 0.8P will again be decreased by 20 %

so 20% of 0.8P = \frac{20}{100} \times 0.8P = (0.2)(0.8P)

So after 2 years value is decreased by (0.2)(0.8P)

so value after 2 years will be = 0.8P - 0.2(0.8P)

taking 0.8P common out we get 0.8P(1-0.2)

= 0.8P(0.8)

=P(0.8)^{2}-------------------------(2)

Similarly after 3 years, this value P(0.8)^{2} will again be decreased by 20 %

so 20% of P(0.8)^{2}  \frac{20}{100} \times P(0.8)^{2} = (0.2)P(0.8)^{2}

So after 3 years value is decreased by (0.2)P(0.8)^{2}

so value after 3 years will be = P(0.8)^{2}   - (0.2)P(0.8)^{2}

taking P(0.8)^{2} common out we get P(0.8)^{2}(1-0.2)

P(0.8)^{2}(0.8)

P(0.8)^{3}-----------------------(3)

so from (1), (2), (3) we can see the following pattern

value after 1 year is P(0.8) or P(0.8)^{1}

value after 2 years is P(0.8)^{2}

value after 3 years is P(0.8)^{3}

so value after x years will be P(0.8)^{x} ( whatever is the year, that is raised to power on 0.8)

So function is best described by exponential model

y = P(0.8)^{x} where y is the value after x years

so thats the final answer

3 0
3 years ago
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