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irina [24]
3 years ago
13

If mints sell three for25°, how many can 1buy with a $20 bill?​

Mathematics
1 answer:
just olya [345]3 years ago
4 0

Answer:

240

Step-by-step explanation:

20/0.25 = 80

80 x 3 = 240

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At 8:00 am the temperature outside was -12oF. By noon, the temperature has risen by 9 oF. What was the temperature at noon? (Jus
Dahasolnce [82]

Answer:

-3

Step-by-step explanation:

It would just be -12+9=-3.

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3 years ago
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You’ve observed the following returns on Crash-n-Burn Computer’s stock over the past five years: 14 percent, –9 percent, 16 perc
MatroZZZ [7]

Answer:

a. 9%

b. 0.01445

c. 12.02%

Step-by-step explanation:

n = 5 years

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A= \frac{14-9+16+21+3}{5} \\A=9.00

b. Historical variance (to 5 decimal places)

V= \frac{\sum (X_i-A)^2}{n-1} \\V= \frac{(0.14-0.09)^2+(-0.09-0.09)^2+(0.16-0.09)^2(0.21-0.09)^2(0.03-0.09)^2}{5-1}\\ V=0.01445

c. Standard deviation.

s = \sqrt{V} \\s= \sqrt{0.01445}\\s=0.1202 = 12.02\%

8 0
3 years ago
3)
Wewaii [24]

Answer:

For covering  1 unit area of the entire playground, the amount of sand required is equal to volume of 3 buckets of sand.

Step-by-step explanation:

Given -

\frac{1}{3} volume of sand in bucket is able to cover \frac{1}{9} area of the entire playground

Thus,

For covering  \frac{1}{9} unit area of the entire playground, the amount of sand required is equal to  \frac{1}{3}  of the total volume of sand in bucket

For covering  1 unit area of the entire playground, the amount of sand required is equal to

\frac{\frac{1}{3} }{\frac{1}{9} } \\\\\frac{1}{3}  * \frac{9}{1} \\\frac{9}{3}\\= 3

For covering  1 unit area of the entire playground, the amount of sand required is equal to volume of 3 buckets of sand.

4 0
3 years ago
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