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3 years ago

Wall-mart had a laptop for $500. Best Buy was selling the same laptop for 15% more. What is price of the laptop at Best Buy?

Mathematics

2 answers:

Allisa [31]3 years ago

8 0

Answer:

575

Step-by-step explanation:

15% of 500 = 75

500 + 75 = 575

bonufazy [111]3 years ago

7 0

Answer:

$425

Step-by-step explanation:

500*0.15=75

500-75=524

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The number of people signed up for a bus trip increased from 32 to 45. What is the percent increase? Round to the nearest percen

Lynna [10]

To calculate the percentage increase: First: work out the difference (increase) between the two numbers you are comparing. Then: divide the increase by the original number and multiply the answer by 100.

45-32 = 13 / 32 *100= 40.625 round to the nearest percent is 41%

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3 years ago

Ice cream cost $4.96 for 4 litres how much for 6 litres

aivan3 [116]

Answer:

$7.44

Step-by-step explanation:

$4.96/ 2= 2.48

$2.48 X 3 = 7.44

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3 years ago

Read 2 more answers

The graph for the equation y=-2x+1 is shown below. If another equation is graphed so that the system has no solution, which equa

12345 [234]

Y=-2x+3 (could be any y-int as long as it has the same slope)

This line is parallel to the original line so they never intersect

Therefore the answer is no solution

5 0

3 years ago

Eli invested $ 330 $330 in an account in the year 1999, and the value has been growing exponentially at a constant rate. The val

Cloud [144]

Answer:

The value of the account in the year 2009 will be $682.

Step-by-step explanation:

The acount's balance, in t years after 1999, can be modeled by the following equation.

$A(t) = Pe^{rt}$

In which A(t) is the amount after t years, P is the initial money deposited, and r is the rate of interest.

$330 in an account in the year 1999

This means that $P = 330$

$590 in the year 2007

2007 is 8 years after 1999, so P(8) = 590.

We use this to find r.

$A(t) = Pe^{rt}$

$590 = 330e^{8r}$

$e^{8r} = \frac{590}{330}$

$e^{8r} = 1.79$

Applying ln to both sides:

$\ln{e^{8r}} = \ln{1.79}$

$8r = \ln{1.79}$

$r = \frac{\ln{1.79}}{8}$

$r = 0.0726$

Determine the value of the account, to the nearest dollar, in the year 2009.

2009 is 10 years after 1999, so this is A(10).

$A(t) = 330e^{0.0726t}$

$A(10) = 330e^{0.0726*10} = 682$

The value of the account in the year 2009 will be $682.

4 0

4 years ago

Michelle has a jar of quarters and dimes totalling to 3.10 if there are sixteen coins in the jar how many more quarters are ther

Sav [38]

Q + d = 16....q = 16 - d
0.25q + 0.10d = 3.10

0.25(16 - d) + 0.10d = 3.10
4 - 0.25d + 0.10d = 3.10
-0.25d + 0.10d = 3.10 - 4
-0.15d = -0.90
d = -0.90/-0.15
d = 6...dimes

q + d = 16
q + 6 = 16
q = 16 - 6
q = 10...quarters

so there are (10 - 6) = 4 more quarters then dimes

5 0

3 years ago