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algol [13]
3 years ago
5

The value given below is discrete. Use the continuity correction and describe the region of the normal distribution that corresp

onds to the indicated probability. Probability of fewer than 8 passengers who do not show up for a flight Choose the correct answer below. A. The area to the right of 8.5 B. The area between 7.5 and 8.5 C. The area to the right of 7.5 D. The area to the left of 7.5 E. The area to the left of 8.5
Mathematics
1 answer:
lions [1.4K]3 years ago
6 0

Answer:

D. The area to the left of 7.5

Step-by-step explanation:

If fewer than 8 passengers do not show up for a flight, then at most 7 passengers do not show up for a flight. Therefore, applying continuity correction, the area under the curve must comprehend from zero to seven passengers not showing up, but not the 8th passenger. Thus, the area must be to the left of 7.5.

The answer is D. The area to the left of 7.5.

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HELP PLEASE ITS DUE TOMMOROW AND I HAVE 2 TESTS TO TAKE TODAY D:
Mashcka [7]

Answer:

x=25

x=47

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Step-by-step explanation:

2x+10=60

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2x=94

2x/2=94/2

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5y-25=100

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5y/5=125/5

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11x=110

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3 years ago
Solve the system of linear equations by elimination x + 2y = 13 --x+y=5​
katen-ka-za [31]

Answer:

{x,y} = {1,6}  Have a great day

5 0
3 years ago
Read 2 more answers
Help!! 50 points and brainliest!
Viktor [21]

Answer:

Second choice:

x=2t

y=4t^2+4t-3

Fifth choice:

x=t+1

y=t^2+4t

Step-by-step explanation:

Let's look at choice 1.

x=t+1

y=t^2+2t

I'm going to subtract 1 on both sides for the first equation giving me x-1=t. I will replace the t in the second equation with this substitution from equation 1.

y=(x-1)^2+2(x-1)

Expand using the distributive property and the identity (u+v)^2=u^2+2uv+v^2:

y=(x^2-2x+1)+(2x-2)

y=x^2+(-2x+2x)+(1-2)

y=x^2+0+-1

y=x^2

So this not the desired result.

Let's look at choice 2.

x=2t

y=4t^2+4t-3

Solve the first equation for t by dividing both sides by 2:

t=\frac{x}{2}.

Let's plug this into equation 2:

y=4(\frac{x}{2})^2+4(\frac{x}{2})-3

y=4(\frac{x^2}{4})+2x-3

y=x^2+2x-3

This is the desired result.

Choice 3:

x=t-3

y=t^2+2t

Solve the first equation for t by adding 3 on both sides:

x+3=t.

Plug into second equation:

y=(x+3)^2+2(x+3)

Expanding using the distributive property and the earlier identity mentioned to expand the binomial square:

y=(x^2+6x+9)+(2x+6)

y=(x^2)+(6x+2x)+(9+6)

y=x^2+8x+15

Not the desired result.

Choice 4:

x=t^2

y=2t-3

I'm going to solve the bottom equation for t since I don't want to deal with square roots.

Add 3 on both sides:

y+3=2t

Divide both sides by 2:

\frac{y+3}{2}=t

Plug into equation 1:

x=(\frac{y+3}{2})^2

This is not the desired result because the y variable will be squared now instead of the x variable.

Choice 5:

x=t+1

y=t^2+4t

Solve the first equation for t by subtracting 1 on both sides:

x-1=t.

Plug into equation 2:

y=(x-1)^2+4(x-1)

Distribute and use the binomial square identity used earlier:

y=(x^2-2x+1)+(4x-4)

y=(x^2)+(-2x+4x)+(1-4)

y=x^2+2x+-3

y=x^2+2x-3.

This is the desired result.

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Answer:

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Step-by-step explanation:

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