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Lady bird [3.3K]
3 years ago
7

If the exercise ball has been filled with Nitrogen which has a density of 0.2 grams per cubic inches. Find the mass in grams.

Mathematics
1 answer:
Jobisdone [24]3 years ago
4 0

Answer:

m = 11581.2 grams

Step-by-step explanation:

Given that,

The density of the ball, d = 0.2 g/in³

The radius of the ball, r = 24 inches

We need to find the mass of the ball in grams.

The density of an object is hiven by :

d=\dfrac{m}{V}\\\\d=\dfrac{m}{\dfrac{4}{3}\pi r^3}\\\\m=\dfrac{4}{3}\pi r^3\times d

Put all the values,

m=\dfrac{4}{3}\pi \times 24^3\times 0.2\\\\m=11581.16\ g

or

m = 11581.2 grams

Hence, the mass of the ball is 11581.2 grams.

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What is the distance between the points (23 , 11) and (4 , 11) in the coordinate plane?
Bas_tet [7]

Answer:just do 23 minus 6 and that should be the answer because the y-ccordinate is in the same place

Step-by-step explanation:

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3 years ago
Four more than seven times a number is 12.” Write an equation that represents the statement. Then solve.
Leona [35]

7x + 4 = 12

7x = 12-4

7x = 8

x = 7/8

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4 years ago
Write two expressions to show a number increased by 11. Then, draw models to prove that both expressions represent the same thin
ioda

Answer:

n + 11 and 11 + n

Step-by-step explanation:

Let n be the unknown number,

n is increased by 11,

That is, n + 11

By the commutative property of addition,

The expression would be,

11 + n

For drawing a model that shows n + 11

Take two boxes in which first shows n and second shows 11 and add them,

Similarly, for showing 11 + n, take first box that shows 11 and second box that shows n then add them.

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4 years ago
When an automobile is stopped with a roving safety patrol,each tire is checked for tire wear, and each headlight is checkedto se
ryzh [129]

Answer:

a) Joint ptobability distribution

\begin{pmatrix}  &Y=0&Y=1&Y=2&Y=3&Y=4\\X=0&0.3&0.05&0.025&0.025&0.1\\ X=1&0.18&0.03&0.015&0.015&0.06 \\ X=2&0.12&0.02&0.01&0.01&0.04\end{pmatrix}

b) P(X<= 1 and Y <= 1) = P(X<= 1) * P(Y<=1) = 0.56

c) P(X + Y = 0)=0.3

d) P(X + Y <= 1)=0.53

Step-by-step explanation:

We have to construct the joint probability table with the marginal probabilities of X and Y.

X can take values from 0 to 2, and Y can take values from 0 to 4.

We can calculate each point of the joint probability as:

P(x,y)=P_x(x)*P_y(y)

Then, the joint probabilities are:

X=0 Y=0 Px=0.5 Px=0.6 P(0,0)=0.3

X=0 Y=1 Px=0.5 Px=0.1 P(0,1)=0.05

X=0 Y=2 Px=0.5 Px=0.05 P(0,2)=0.025

X=0 Y=3 Px=0.5 Px=0.05 P(0,3)=0.025

X=0 Y=4 Px=0.5 Px=0.2 P(0,4)=0.1

X=1 Y=0 Px=0.3 Px=0.6 P(1,0)=0.18

X=1 Y=1 Px=0.3 Px=0.1 P(1,1)=0.03

X=1 Y=2 Px=0.3 Px=0.05 P(1,2)=0.015

X=1 Y=3 Px=0.3 Px=0.05 P(1,3)=0.015

X=1 Y=4 Px=0.3 Px=0.2 P(1,4)=0.06

X=2 Y=0 Px=0.2 Px=0.6 P(2,0)=0.12

X=2 Y=1 Px=0.2 Px=0.1 P(2,1)=0.02

X=2 Y=2 Px=0.2 Px=0.05 P(2,2)=0.01

X=2 Y=3 Px=0.2 Px=0.05 P(2,3)=0.01

X=2 Y=4 Px=0.2 Px=0.2 P(2,4)=0.04

We can write it in the form of a matrix:

\begin{pmatrix}  &Y=0&Y=1&Y=2&Y=3&Y=4\\X=0&0.3&0.05&0.025&0.025&0.1\\ X=1&0.18&0.03&0.015&0.015&0.06 \\ X=2&0.12&0.02&0.01&0.01&0.04\end{pmatrix}

b) From the joint probability P(X<= 1 and Y <= 1) is equal to

P(X\leq 1 \& Y \leq 1)=P(0,0)+P(0,1)+P(1,0)+P(1,1)\\\\P(X\leq 1 \& Y \leq 1)=0.3+0.05+0.18+0.03=0.56

We can calculate P(X<= 1) * P(Y<=1)

P_x(X\leq 1)=P_x(0)+P_x(1)=0.5+0.3=0.8\\\\P_y(Y\leq1)=P_y(0)+P_y(1)=0.6+0.1=0.7\\\\P_x(X\leq1)*P_y(Y\leq1)=0.8*0.7=0.56

Both calculations give the same result.

c) Probability of no violations

P(X+Y=0)=P(0,0)=0.3

d) P(X + Y <= 1)

P(X+Y \leq 1)=P(0,0)+P(0,1)+P(1,0)\\\\P(X+Y \leq 1)=0.3+0.05+0.18=0.53

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4 years ago
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