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Ludmilka [50]
3 years ago
7

What is the interquartile range of the data? 150, 165, 180, 198, 204, 216, 230, 255 A. 105 B. 51 C. 50.5 D. 28.5 ASAP!!! PLEASE

ASAP
Mathematics
1 answer:
Reika [66]3 years ago
7 0
<span>I Think the answer is a.

</span>
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Mariah's earnings rate can be described in dollars per hour or hours per dollar. Mariah earns $10 for each hour she works at a m
weeeeeb [17]

Answer:

<u>dollars per hour rate = $10</u>

<u>hours per dollar = 6 minutes 15 seconds</u>

Step-by-step explanation:

Take note that one is singular (dollar) without an 's', and the other is plural (dollars). Thus, since her dollars per hour rate tells us how much she will earn for each hour of work, we would expect her hours per dollar rate to tell us how much she has to work in minutes to earn a single dollar ($1).

If we divide \frac{10}{60 minutes} = $0.16 which indicates how much she earns for every minute she works at the theater. Further dividing this value into 1 minute we find the hour per dollar rate 1/0.16 = 6.25. minutes.

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3 years ago
I don't know help please
Iteru [2.4K]

Answer:

huh

Step-by-step explanation:

did you add anything?

5 0
3 years ago
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What is the answer to 8x + 15 = 3m + 5
Naddika [18.5K]
Solving for x
x=3/8m-5/4,m



Solving for m
m=10/3+8/3x,x
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4 years ago
jeremy mowed several lawns to earn money for camp. after he paid $17 for gas, he had $75 leftover to pay towards camp. write and
USPshnik [31]
Money earned = e=75+17=92
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3 years ago
The desired percentage of sio2 in a certain type of aluminous cement is 5.5. to test whether the true average percentage is 5.5
LekaFEV [45]
Given that t<span>he desired percentage of sio2 in a certain type of aluminous cement is 5.5. to test whether the true average percentage is 5.5 for a particular production facility, 16 independently obtained samples are analyzed. suppose that the percentage of sio2 in a sample is normally distributed with σ = 0.32 and that \bar{x}=5.24.

</span>
<span>To investigate whether this indicate conclusively that the true average percentage differs from 5.5.



Part A:

From the question, it is claimed that </span><span>t<span>he desired average percentage of sio2 in a certain type of aluminous cement is 5.5</span></span> and we want to test whether the information from the random sample <span>indicate conclusively that the true average percentage differs from 5.5.

Therefore, the null hypothesis and the alternative hypothesis is given by:

H_0:\mu=5.5 \\  \\ H_a:\mu\neq5.5



Part B:

The test statistics is given by:

z= \frac{\bar{x}-\mu}{\sigma/\sqrt{n}}  \\  \\ =\frac{5.25-5.5}{0.32/\sqrt{16}} \\  \\ = \frac{-0.25}{0.32/4} = -\frac{0.25}{0.08}  \\  \\ =-3.125



Part C:

The p-value is given by

P(z\ \textless \ -3.125)=1-P(z



Part D:

Because the p-value is less than the significant level α, we reject the null hypothesis and conclude that "</span><span>There is sufficient evidence to conclude that the true average percentage differs from the desired percentage."



Part E:

</span>If the true average percentage is μ = 5.6 and a level α = 0.01 test based on n = 16 is used, what is the probability of detecting this departure from H0? (Round your answer to four decimal places.)

The probability of detecting the departure from H_0 is given by

1-\phi\left(z_{1-\frac{\alpha}{2}}+ \frac{\mu_0-\mu_1}{\sigma/\sqrt{n}} \right)+\phi\left(-z_{1-\frac{\alpha}{2}}+ \frac{\mu_0-\mu_1}{\sigma/\sqrt{n}} \right) \\  \\ =1-\phi\left(z_{1-\frac{0.01}{2}}+ \frac{5.5-5.6}{0.32/\sqrt{16}} \right)+\phi\left(-z_{1-\frac{0.01}{2}}+ \frac{5.5-5.6}{0.32/\sqrt{16}} \right) \\  \\ =1-\phi\left(z_{1-0.005}+ \frac{-0.1}{0.32/4} \right)+\phi\left(-z_{1-0.005}+ \frac{-0.1}{0.32/4} \right)

=1-\phi\left(z_{0.995}+ \frac{-0.1}{0.08} \right)+\phi\left(-z_{0.995}+ \frac{-0.1}{0.08} \right) \\  \\ =1-\phi(2.576-1.25)+\phi(-2.576-1.25) \\  \\ =1-\phi(1.326)+\phi(-3.826) \\  \\ =1-0.90758+0.00007 \\  \\ =0.0925



Part F:

What value of n is required to satisfy α = 0.01 and β(5.6) = 0.01? (Round your answer up to the next whole number.)

The value of n is required to satisfy α = 0.01 and β(5.6) = 0.01 is given by

n=\left[ \frac{\sigma(z_{0.005}+z_{0.01})}{\mu_0-\mu} \right]^2 \\  \\ = \left[\frac{0.32(-2.576-2.326)}{5.5-5.6} \right]^2 \\  \\ =\left[\frac{0.32(-4.902)}{-0.1} \right]^2=\left[\frac{-1.56864}{-0.1} \right]^2 \\  \\ =(15.6864)^2=247
3 0
4 years ago
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