A soft drink-dispensing machine uses paper cups that hold a maximum of 12 ounces. The machine is set to dispense a mean of 10 ounces of the drink. Because of an imprecise process, the amount dispensed actually varies; it is normally distributed with a standard deviation of 1 ounce. If we simulated the filling process, and had the random number .564, how many ounces would the cup contain? Give your answer to 3 decimal places.

The cup would contain 10.16 ounces.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Mean of 10 ounces, standard deviation of 1 ounce:

This means that

If we simulated the filling process, and had the random number .564, how many ounces would the cup contain?

This means that we have to find X when Z has a pvalue of 0.564. So X when Z = 0.16. So

The cup would contain 10.16 ounces.

For 1 flavor there are 9 topping

Therefore, for 5 different flavors there will be 5*9 choices

No of choices= 5*9

=45 

The answer is in the image 

The answer is in the image 

y=2x+15

where y=Value of coin

x=Age in years

Value of coin after 19 years=2*19+15

=$53

Therefore, Value after 19 years=$53

The answer is in the image 

F=ma

where F=force

m=mass

a=acceleration

Here,

F=4300

a=3.3m/s2

m=F/a

    =4300/3.3

    =1303.03kg

F=ma

where F=force

m=mass

a=acceleration

Here,

F=4300

a=3.3m/s2

m=F/a

    =4300/3.3

    =1303.03kg

Approximately it is aqual to 1300kg

The answer is in the image