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marissa [1.9K]
4 years ago
9

Marlene lives in a building where half the residents live on the first floor and half live on the second floor. She wants to est

imate the probability that, out of 3 randomly selected residents, 1 or 2 live on the first floor.
She uses a coin to conduct a simulation, by letting H represent someone living on the first floor and T represent someone living on the second floor and then flipping the coin three times. She repeats this process for a total of 15 trials. The results are shown in the table.

Estimated probability that 1 or 2 of 3 randomly selected residents live on the first floor: ________

A.1/2
B.1/6
C.5/6
D.5/12

Mathematics
2 answers:
ludmilkaskok [199]4 years ago
8 0
The correct answer is:       5/6
Liula [17]4 years ago
7 0
5/6........................................................
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When Lorenzo finished his exam last week, he thought the test was over. But the instructor put z-scores on each student's paper
Anarel [89]

Answer:

Lorenzo's score on exam was 75.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 61

Standard Deviation, σ = 8

We are given that the distribution of test score is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

z-score = 1.75

We have to find the value of x.

Putting values, we get

1.75 = \dfrac{x - 61}{8}\\\\x = 1.75(8) + 61\\x = 75

Thus, Lorenzo's score on exam was 75.

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3 years ago
Jonathan's parents told him that for every 5 hours of Academic Prep work he completes, he would be able to play 3 hours of video
Over [174]

Answer: Jonathan spends more time playing video games because 3/5 of the time he is playing video games, while Lucas is only playing thirty minutes out of sixty, which simplifies into 1/2. Jonathan plays 60% of the time and Lucas plays 50%.

Step-by-step explanation:

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3 years ago
Multiple-Choice Integration, Picture Included, Please Include Work
ValentinkaMS [17]
\displaystyle\int_0^2\sqrt{4-x^2}\,\mathrm dx

Recall that a circle of radius 2 centered at the origin has equation

x^2+y^2=4\implies y=\pm\sqrt{4-x^2}

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You have r=2, so the definite integral is equal to \dfrac{2^2\pi}4=\pi.

Another way to verify this is to actually compute the integral. Let x=2\sin u, so that \mathrm dx=2\cos u\,\mathrm du. Now

\displaystyle\int_0^2\sqrt{4-x^2}\,\mathrm dx=\int_0^{\pi/2}\sqrt{4-(2\sin u)^2}(2\cos u)\,\mathrm du=4\int_0^{\pi/2}\cos^2u\,\mathrm du

Recall the half-angle identity for cosine:

\cos^2u=\dfrac{1+\cos2u}2

This means the integral is equivalent to

\displaystyle2\int_0^{\pi/2}(1+\cos 2u)\,\mathrm du=2u+\sin2u\bigg|_{u=0}^{u=\pi/2}=\pi
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10 squared plus 18 squared equals C squared is the set up
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A customer went to a garden shop and bought some potting soil for $11.50 and 9 shrubs. The total bill was $94.75. Write and solv
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So the answer to this question is:

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