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earnstyle [38]
2 years ago
12

Which of the following are factors of P ( x ) = 2 x 3 + 3 x 2 − 8 x − 12 ?​

Mathematics
2 answers:
alukav5142 [94]2 years ago
6 0

Answer:

108

Step-by-step explanation:

12-8(-12)

12-(-96)

12+96

108

prisoha [69]2 years ago
4 0
108 is the answer to the following
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Peter attends 6 dance lessons each week all year long a year has 52 weeks peter missed 5 dance lessons while sick how many dance
viktelen [127]
6 lessons per week multiplied by the number of weeks in a year
6 x 52 = 312
total - 5 missed
312 - 5 = 307 dance lessons
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3 years ago
Square root of 476 is?
steposvetlana [31]

Answer:

The square root of 476 is 21.8

5 0
3 years ago
Read 2 more answers
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
Read 2 more answers
Multiplication question
Gemiola [76]
Well one is 5 times 5
7 0
3 years ago
Find a number that is between2/9and3/11
SVETLANKA909090 [29]
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Change the denominators to be the same:
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\text {Between } \dfrac{22}{99}  \text { and }  \dfrac{27}{99}

\text {Possible answers are  } \dfrac{23}{99} \ , \dfrac{24}{99} \ , \dfrac{25}{99} \text { and }  \dfrac{26}{99}

3 0
3 years ago
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