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Oksanka [162]
2 years ago
9

Y > − 3/4 x + 3 please help​

Mathematics
2 answers:
pshichka [43]2 years ago
4 0

Answer:

D

Step-by-step explanation:

sweet-ann [11.9K]2 years ago
3 0

Answer:

C im pretty sure, could be wrong.

Step-by-step explanation:

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Solve: |v + 8| − 5 = 2
AveGali [126]

Answer:

v¹= -15, v²= -1

Step-by-step explanation:

there's 2 solutions to this

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Which point represents 2 1/4 on the graph
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hope this helps

Step-by-step explanation:

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A cola-dispensing machine is set to dispense 8 ounces of cola per cup, with a standard deviation of 1.0 ounce. The manufacturer
pshichka [43]

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} } }

X[bar]= 7.72ounces

The lower control point will be 7.72 ounces.

The upper value will be:

Z_{0.95}= 1.648

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

X[bar]= 8.28ounces

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)

P(Z≤(8.28-7.6)/(1/√34))- P(Z≤7.72-7.6)/(1/√34))

P(Z≤7.11)- P(Z≤0.70)= 1 - 0.758= 0.242

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!

5 0
3 years ago
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You pick a card at random.
slega [8]
25%

There are 4 different numbers you could draw. Having all 4 cards would be equal to 100%, making each card worth 25%.
25 + 25 + 25 + 25= 100
6 0
3 years ago
PLEASE HELP ASAP!!
Art [367]

Answer:

See explanation

Step-by-step explanation:

A.  P(top)=top outcomes/all kinds of outcomes=4/30=2/15=13.33333333333...%

P(bottom)=bottom outcomes/ all kinds of outcomes=1/30=3.33333333....%

P(side)=25/30=5/6=83.33333333333...%

B. No. If were equally likely, the probabilities for A would have been roughly the same. It seems like the side event is more probable.

7 0
3 years ago
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