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german
3 years ago
11

A woman who weighs 160 pounds on earth weighs 30 on the moon. If a boy weighs 60 pounds on earth , how much dose he weigh on the

moon.
Mathematics
2 answers:
soldi70 [24.7K]3 years ago
6 0

Answer:

10 pounds

Step-by-step explanation:

Igoryamba3 years ago
3 0
60Kg. Mass does not vary by the strength of the gravitational field. What changes is the weight which will only be about one-sixth of what it is on Earth. So a 60Kg mass on the Moon would weigh approximately the same as a 10Kg mass on the Earth.
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Kendra swims 126 miles in 3 days, what is the constant rate that Kendra swims per day? a. 40 miles per day c. 42 miles per day b
Leni [432]

Answer:

c. 42 miles per day

Step-by-step explanation:

From the information given, you can use a rule of three to calculate the constant rate that Kendra swims per day given that she swims 126 miles in 3 days:

3 days → 126 miles

1 day   →        x

x=(1*126)/3=42 miles

According to this, the answer is that Kendra swims 42 miles per day.

5 0
3 years ago
What is 5/7 in the simplest form
Alexeev081 [22]
Exact form: 
5/7

Decimal form:
0.71428571... 


3 0
3 years ago
Sierra has plotted two vertices of a rectangle at (3,2) and (8,2). What is the length of the sides of the rectangle
Eddi Din [679]

Answer:

5units

Step-by-step explanation:

Given the two vertices of a rectangle at (3,2) and (8,2). We are to calculate the distance between the two points.

Using the formula

D = √(x2-x1)²+(y2-y1)²

D = √(2-2)²+(8-3)²

D = √0+5²

D =  √25

D = 5

Hence the length of the side of the rectangle is 5units

4 0
3 years ago
Oliver has a box of toy cars, but he is not sure how many he has. If he arranges the cars in groups of four, he has three left o
Zarrin [17]
X/4 = r3 or 3/4
x/3 = r2 or 2/3
x/5 = r3 or 3/5

we know x is not divisible by 4 3 or 5 and is less than 30.

we know the value ends in 3 or 8 from x/5 = r3

we know the value is greater than 3 from r3

so far we got 13 and 23

13 is eliminated from x/4 = r3 because it is r1 not r3

so the answer is 23




7 0
3 years ago
The number of people in a car that crosses a certain bridge is represented by the random variable X, which has a mean value μX =
Alona [7]

Let Y be the total amount of money paid by any given set of passengers. If there are X passengers in a car, then the driver must pay a toll of Y=0.5X+3.

Then Y has first moment (equal to the mean)

E[Y]=E[0.5X+3]=0.5E[X]+3E[1]=0.5\mu_X+3=\boxed{4.35}

and second moment

E[Y^2]=E[0.25X^2+3X+9]=0.25E[X^2]+3E[X]+9E[1]=0.25E[X^2]+3\mu_X+9

Recall that the variance is the difference between the first two moments:

\mathrm{Var}[X]=E[X^2]-E[X]^2\implies E[X^2]={\sigma^2}_X+{\mu_X}^2

\implies E[Y^2]=0.25({\sigma^2}_X+{\mu_X}^2)+3\mu_X+9\approx19.22

\implies\mathrm{Var}[Y]=E[Y^2]-E[Y]^2=\boxed{0.3}

3 0
4 years ago
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