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3 years ago

A rectangular prism is 3 units high, 2 units wide, and 5 units long. What is its surface area

Mathematics

1 answer:

andrew11 [14]3 years ago

8 0

Answer:

62 square units

Step-by-step explanation:

Surface area of a rectangular prism = 2 ( wl + hl + hw)

Where,

w = width = 2 units

h = height = 3 units

l = length = 5 units

Surface area of a rectangular prism = 2 ( wl + hl + hw)

= 2 (2*5 + 3*5 + 3*2)

= 2 (10 + 15 + 6)

= 2 (31)

= 62 square units

Surface area of the rectangular prism = 62 square units

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Solve quadratic equation (x-3)^2+5=14

laiz [17]

Answer:

the Answer is 14=5+2^(3-x)

Step-by-step explanation:

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3 years ago

What is the value of a

MrMuchimi

1. thats because in the alphabet where a=1 and b=2 and c=3, A=1

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3 years ago

a country's population in 1990 was 134 million. in 2000 it was a 139 million. estimate the population in 2014 using the exponent

yuradex [85]

F=ir^t

139=134r^10

139/134=r^10

r=(139/134)^(1/10) then:

f=134(139/134)^(t/10) so in 2014, t=24 so

f=134(139/134)^(2.4)

f≈146 million (to nearest million)

Some will say that you have to use the exponential function, but it really gives you the same answer...even for continuous compounding :)...

A=Pe^(kt)

139=134e^(10k)

139/134=e^(10k)

ln(139/134)=10k

k=ln(139/134)/10 so

A=134e^(t*ln(139/134)/10) when t=24

A=134e^(2.4*ln(139/134))

A≈146 million (to nearest million)

The only real reason or advantage to using A=Pe^(kt) is when you start getting into differential equations...

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3 years ago

Assume that the probability of a defective computer component is 0.02. Components arerandomly selected. Find the probability tha

solmaris [256]

Answer:

0.0177 = 1.77% probability that the first defect is caused by the seventh component tested.

The expected number of components tested before a defective component is found is 50, with a variance of 0.0208.

Step-by-step explanation:

Assume that the probability of a defective computer component is 0.02. Components are randomly selected. Find the probability that the first defect is caused by the seventh component tested.

First six not defective, each with 0.98 probability.

7th defective, with 0.02 probability. So

$p = (0.98)^6*0.02 = 0.0177$

0.0177 = 1.77% probability that the first defect is caused by the seventh component tested.

Find the expected number and variance of the number of components tested before a defective component is found.

Inverse binomial distribution, with $p = 0.02$

Expected number before 1 defective(n = 1). So

$E = \frac{n}{p} = \frac{1}{0.02} = 50$

Variance is:

$V = \frac{np}{(1-p)^2} = \frac{0.02}{(1-0.02)^2} = 0.0208$

The expected number of components tested before a defective component is found is 50, with a variance of 0.0208.

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2 years ago

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Ronch [10]

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3 years ago