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eduard
2 years ago
13

jack wants to paint his walls. two walls are 12 ft by 8 ft and two walls are 10 ft by 8 ft. what is the area to be painted? if o

ne can of paint covers 30 square ft, how many cans are needed?
Mathematics
1 answer:
Ksivusya [100]2 years ago
4 0

Answer:

352 ft (squared)  , he will need 12 paint cans

Step-by-step explanation:

Two 12 by 8ft walls = 192 ft in area, Two 10 by 8ft walls = 160 ft in Area, that equals 352 ft total, 352/30=11.73 repeated, round up to 12

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Do (10 x 10) then (10 x 20), and then add your two answers together.
Its simple :)
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3 years ago
Quiz 10-1 intro circles central angles question 9
kompoz [17]

Answer:

if the diameter of the circle is 25, then the radius will be 12.5.

A=3.14(12.5)2 --> 490.9m^2

C= 2*3.14(12.5)---> 78.5 m

3 0
1 year ago
The table below shows pairs of value that satisfy a linear function.
neonofarm [45]

Answer:

-27

Step-by-step explanation:

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3 years ago
Assume that a procedure yields a binomial distribution with a trial repeated n = 5 times. Use some form of technology to find th
Digiron [165]

Answer:

P(X = 0) = 0.0263

P(X = 1) = 0.1407

P(X = 2) = 0.3012

P(X = 3) = 0.3224

P(X = 4) = 0.1725

P(X = 5) = 0.0369

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

In this problem we have that:

n = 5, p = 0.517

Distribution

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{5,0}.(0.517)^{0}.(0.483)^{5} = 0.0263

P(X = 1) = C_{5,1}.(0.517)^{1}.(0.483)^{4} = 0.1407

P(X = 2) = C_{5,2}.(0.517)^{2}.(0.483)^{3} = 0.3012

P(X = 3) = C_{5,3}.(0.517)^{3}.(0.483)^{2} = 0.3224

P(X = 4) = C_{5,4}.(0.517)^{4}.(0.483)^{1} = 0.1725

P(X = 5) = C_{5,5}.(0.517)^{5}.(0.483)^{0} = 0.0369

6 0
3 years ago
what effect does doubling the radius of a cone have on the volume of the cone if the height remains the same
Butoxors [25]
V=(1/3)hpir^2
lets say this is original, undoubled volume
so o=v=(1/3)hpir^2

so for new volume, or n, that is r to 2r, doubled radius
n=(1/3)hpi(2r)^2
n=(1/3)hpi4r^2
n=4((1/3)hpir^2)
remmember that o=(1/3)hpir^2
n=4(o)

it is 4 times the old one
4 0
3 years ago
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