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3 years ago

Which of the following statements is not true about underage drinking?

Engineering

1 answer:

Natali [406]3 years ago

5 0

Answer:

underage drinking isnt okay

Explanation:

just used the info i have

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a metal coin has certain properties that can be measured.which property of a coin is different on the moon that is on earth?

Sloan [31]

Answer:

Coins weigh less on the Moon.

Explanation:

Gravity is only 1/6th as strong on the Moon than it is on Earth. Where a nickle is about 5 grams on Earth, it is less than 1 gram on the Moon. Gravity is affected by the size of the planet or moon. The Moon is much less massive than the Earth.

8 0

3 years ago

we wish to send at a rate of 10Mbits/s over a passband channel. Assuming that an excess bandwidth of 50% is used, how much bandw

gayaneshka [121]

Answer:

QPSK: 7.5 MHz

64-QAM:2.5 MHz

64-Walsh-Hadamard: 160 MHz

Explanation:

See attached picture.

6 0

3 years ago

Using Pascal’s Law and a hydraulic jack, you want to lift a 4,000 lbm rock. The large cylinder has a diameter of 6 inches.

jolli1 [7]

Answer:

a diameter of D₂ = 0.183 inches would be required

Explanation:

appyling pascal's law

P applied to the hydraulic jack = P required to lift the rock

F₁*A₁ = F₂*A₂

since A₁= π*D₁²/4 , A₂= π*D₂²/4

F₁*π*D₁²/4 = F₂* π*D₂²/4

F₁*D₁²=F₂*D₂²

D₂ = D₁ *√(F₁/F₂)

replacing values

D₂ = D₁ *√(F₁/F₂) = 6 in * √(120 lbf/(4000 lbm * 32.174 (lbf/lbm)) = 0.183 inches

6 0

3 years ago

In a semiconductor manufacturing process, three wafers from a lot are tested. Each wafer is classified as pass or fail. Assume t

gladu [14]

Answer:

F(x) = 0 ; x < 0

0.064 ; 0 ≤ x < 1

0.352 ; 1 ≤ x < 2

0.784 ; 2 ≤ x < 3

1 ; x ≥ 3

Explanation:

Each wafer is classified as pass or fail.

The wafers are independent.

Then, we can modelate X : ''Number of wafers that pass the test'' as a Binomial random variable.

X ~ Bi(n,p)

Where n = 3 and p = 0.6 is the success probability

The probatility function is given by :

$P(X=x)=f(x)=nCx.p^{x}.(1-p)^{n-x}$

Where $nCx$ is the combinatorial number

$nCx=\frac{n!}{x!(n-x)!}$

Let's calculate f(x) :

$f(0)=3C0.(0.6^{0}).(0.4^{3})=0.4^{3}=0.064$

$f(1)=3C1.(0.6^{1}).(0.4^{2})=0.288$

$f(2)=3C2.(0.6^{2}).(0.4^{1})=0.432$

$f(3)=3C3.(0.6^{3}).(0.4^{0})=0.6^{3}=0.216$

For the cumulative distribution function that we are looking for :

$P(X\leq x)=F(x)$

$F(0)=f(0)\\F(1)=f(0)+f(1)\\F(2)=f(0)+f(1)+f(2)\\F(3)=f(0)+f(1)+f(2)+f(3)=1$

$F(0)=0.064\\F(1)=0.064+0.288=0.352\\F(2)=0.064+0.288+0.432=0.784\\F(3)=0.064+0.288+0.432+0.216=1$

The cumulative distribution function for X is :

F(x) = 0 ; x < 0

0.064 ; 0 ≤ x < 1

0.352 ; 1 ≤ x < 2

0.784 ; 2 ≤ x < 3

1 ; x ≥ 3

5 0

3 years ago

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Nataly_w [17]

Outdoor advertising

8 0

4 years ago