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GrogVix [38]
2 years ago
12

Given that BD ~ BC, find m A. 54 B. 57 C. 64 D. 66

Mathematics
2 answers:
belka [17]2 years ago
3 0

Answer:

b

Step-by-step explanation:

raketka [301]2 years ago
3 0

Answer:

B)  57

Step-by-step explanation:

since BD ≅ BC then ∡C = ∡D

6x-9 = 3x + 24

subtract 3x from each side to get:

3x-9 = 24

add 9 to each side to get:

3x = 33

divide each side by 3 to get:

x = 11

substitute 11 into 'x' in 3x + 24 to get:

3(11) + 24 = 33+24 = 57

check:

substitute 11 into 'x' in 6x - 9 to get:

6(11) - 9 = 66 - 9 = 57

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Luisa had $29.90 in her piggy bank. She made $5.50 a week for 4 weeks by baby-sitting. Write a sequence to show how much money s
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3 years ago
A cola-dispensing machine is set to dispense 8 ounces of cola per cup, with a standard deviation of 1.0 ounce. The manufacturer
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Answer:

Step-by-step explanation:

Hello!

The variable of interest is X: ounces per cup dispensed by the cola-dispensing machine.

The population mean is known to be μ= 8 ounces and its standard deviation σ= 1.0 ounce. Assuming the variable has a normal distribution.

A sample of 34 cups was taken:

a. You need to calculate the Z-values corresponding to the top 5% of the distribution and the lower 5% of it. This means you have to look for both Z-values that separates two tails of 5% each from the body of the distribution:

The lower value will be:

Z_{o.o5}= -1.648

You reverse the standardization using the formula Z= \frac{X[bar]-Mu}{\frac{Sigma}{\sqrt{n} } } ~N(0;1)

-1.648= \frac{X[bar]-8}{\frac{1}{\sqrt{34} } }

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The upper value will be:

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The upper control point will be 8.82 ounces.

b. Now μ= 7.6, considering the control limits of a.

P(7.72≤X[bar]≤8.28)= P(X[bar]≤8.28)- P(X[bar]≤7.72)

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There is a 0.242 probability of the sample means being between the control limits, this means that they will be outside the limits with a probability of 1 - 0.242= 0.758, meaning that the probability of the change of population mean being detected is 0.758.

b. For this item μ= 8.7, the control limits do not change:

P(7.72≤X[bar]≤8.28)= P(X[bar]≤8.28)- P(X[bar]≤7.72)

P(Z≤(8.28-8.7)/(1/√34))- P(Z≤7.72-8.7)/(1/√34))

P(Z≤-2.45)- P(Z≤-5.71)=0.007 - 0= 0.007

There is a 0.007 probability of not detecting the mean change, which means that you can detect it with a probability of 0.993.

I hope it helps!

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