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3 years ago

7

Last weekend, Raptor Club members spotted 160 hawks. This weekend they saw 70% as many as they saw last weekend. How many hawks

did they see this weekend?

1 answer:

pickupchik [31]3 years ago

4 0

They have seen 112 Hawks

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Kay inherits $250,000 which she invests today at a rate of return of 9% compounded annually. How much will Kay's investment be w

SIZIF [17.4K]

Answer: $272,500

I hit this by first doing the step by step things and then adding all of them up.

6 0

3 years ago

Jennifer has a credit card with an APR of 10.22% and a billing cycle of 30 days. The following table shows Jennifer’s credit car

Nimfa-mama [501]

Answer:

The answer is c.

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

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

Please factorise 16p²-9q² for me<br>

kirill115 [55]

Answer:

a ^ 2 - b ^ 2 = ( a + b ) ( a - b ) where a = 4p and b = 3q.

( 4p + 3q ) ( 4p - 3q )

7 0

3 years ago

Shannon collects rainwater to water her flowers. She has one bucket with 3.4 gallons of water and another with 1.85 gallons less

lukranit [14]

Answer:

3.4-1.85 = 1.55 3.4+1.55=4.95

Step-by-step explanation:

5 0

3 years ago