An ice cream cone is filled with vanilla and chocolate ice cream at a ratio of 2:1. If the diameter of the cone is 2 inches and the height is 6 inches, approximately what is the volume of vanilla ice cream in the cone? (round to nearest tenth) A) 1.0 in3 B) 2.1 in3 C) 4.2 in3 D) 6.3 in3

Answer and explanation:

Given : The probabilities of poor print quality given no printer problem, misaligned paper, high ink viscosity, or printer-head debris are 0, 0.3, 0.4, and 0.6, respectively.

The probabilities of no printer problem, misaligned paper, high ink viscosity, or printer-head debris are 0.8, 0.02, 0.08, and 0.1, respectively.

Let the event E denote the poor print quality.

Let the event A be the no printer problem i.e. P(A)=0.8

Let the event B be the misaligned paper i.e. P(B)=0.02

Let the event C be the high ink viscosity i.e. P(C)=0.08

Let the event D be the printer-head debris i.e. P(D)=0.1

and the probabilities of poor print quality given printers are

First we calculate the probability that print quality is poor,

a. Determine the probability of high ink viscosity given poor print quality.

b. Given poor print quality, what problem is most likely?

Probability of no printer problem given poor quality is

Probability of misaligned paper given poor quality is

Probability of printer-head debris given poor quality is

From the above conditional probabilities,

The printer-head debris problem is most likely given that print quality is poor.

Answer:

Answer of Part(a) is 16/49

and Answer of Part(b) is Printer-head debris

Step-by-step explanation:

Answer is in the following attachment

Answer:

The 98 percent confidence interval for the population mean is between 27.85 and 29.75 subjects.

Step-by-step explanation:

We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Ztable as such z has a pvalue of .

So it is z with a pvalue of , so

Now, find M as such

In which is the standard deviation of the population and n is the size of the sample.

The lower end of the interval is the mean subtracted by M. So it is 28.8 - 0.9488 = 27.85 subjects.

The upper end of the interval is the mean added to M. So it is 28.8 + 0.9488 = 29.75 subjects.

The 98 percent confidence interval for the population mean is between 27.85 and 29.75 subjects.

Answer:

x = 7

Step-by-step explanation:

I hope this helps!

Answer:

option (C) 173.8 mg less than < mu μ less than <196.2 mg

Step-by-step explanation:

Data provided ;

number of sample, n = 12

Mean = 185 milligram

standard deviation, s = 17.6 milligrams

confidence level = 95%

α = 0.05                                     [for 95% confidence level]

df = n - 1 = 12 - 1 = 11

Now,

Confidence interval = Mean ± E

here,

E is the margin of error =

also,

=    

=   2.201                         [ from standard t value table]

Thus,

E =

or

E = 11.182 milligrams ≈ 11.2 milligrams

Therefore,

Confidence interval:

Mean - E < μ < Mean + E

or

185 - 11.2 < μ < 185 + 11.2

or

173.8 < μ < 196.2

Hence,

the correct answer is option (C) 173.8 mg less than < mu μ less than <196.2 mg

Final answer:

To construct a confidenceinterval for the population mean cholesterol content of all chicken eggs with a 95% confidence level, we use the sample mean, standard deviation, and sample size to calculate the margin of error. The confidence interval is then constructed by subtracting the margin of error from the sample mean and adding it to the sample mean.

Explanation:

To construct a confidenceinterval for the population mean cholesterol content of all chicken eggs, we first need to find the margin of error. The margin of error depends on the samplemean, standard deviation, sample size, and the desired level of confidence. In this case, we have a sample mean of 185 mg, a standard deviation of 17.6 mg, and a sample size of 12. Since we want a 95% confidence interval, we use a z-score of 1.96. The margin of error is then calculated as 1.96 * (17.6/sqrt(12)), which is approximately 9.61 mg. We can then construct the confidenceinterval by subtracting the margin of error from the sample mean and adding it to the sample mean. Therefore, the 95% confidence interval for the true mean cholesterol content of all such eggs is 175.9 mg to 194.1 mg.

Learn more about Constructing confidence intervals here:

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