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3 years ago

15

Mike and his best friend found some money under the couch. they split the money evenly, each getting $6.02. how much money did t

hey find?

1 answer:

svp [43]3 years ago

7 0

they found $12.04 under the couch and split into two ways getting $6.02 each

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the length of a rectangular parking lot at the airport is 2/3 is mile.if the area is 1/2 square mile,what is the width of the pa

ch4aika [34]

Answer:

3/4

Step-by-step explanation:

3/4*2/3+1/2

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2 years ago

Estimate the temperature at which the rate of chirping is 130 per minute

Andrej [43]

Answer:

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3 years ago

If two points on a line are A(10, −3) and B(12, 9), the rise is __________, and the run is __________, so the slope of the line

kherson [118]

The <u>rise</u> is the difference in y-coordinates:

y₂ - y₁ = 9 - (-3) = 12

The <u>run</u> is the difference in x-coordinates:

x₂ - x₁ = 12 - 10 = 2

The <u>slope</u> is the quotient of rise over run:

m = 12/2 = 6

========================

If two points on a line are A(10, −3) and B(12, 9), the rise is <u>12</u>, and the run is <u>2</u>, so the slope of the line is <u>6</u>.

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2 years ago

What is the metric equivalent in meters to 1250 feet?

Dvinal [7]

381 meters because you divide 1250 by 3.281 and you get 381 meters

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3 years ago

Read 2 more answers

If the mean weight of 4 backfield members on the football team is 221 lb and the mean weight of the 7 other players is 202 lb, w

MatroZZZ [7]

Let $x_1, x_2, x_3, x_4, x_5, x_6, x_7, x_8, x_9, x_{10}, x_{11}$ be the weight of i-th player.

1. If the mean weight of 4 backfield members on the football team is 221 lb, then

$\dfrac{x_1+x_2+x_3+x_4}{4}=221\ lb.$

2. If the mean weight of the 7 other players is 202 lb, then

$\dfrac{x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}}{7}=202\ lb.$

3. From the previous statements you have that

$x_1+x_2+x_3+x_4=221\cdot 4=884 \lb,\\ \\x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}=202\cdot 7=1414\ lb.$

Add these two equalities and then divide by 11:

$x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}=884+1414=2298\ lb,\\ \\\dfrac{x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}}{11}=\dfrac{2298}{11}=208\dfrac{10}{11}\ lb.$

Answer: the mean weight of the 11-person team is $208\dfrac{10}{11}\ lb.$

4 0

2 years ago