julie buys 2kg of apples and 7kg of pears £12.70. the pears cost £1.30 per kilogram what is the price per kilogram of the apples

Answer: C. Significant at 0.036

Step-by-step explanation:

Given:

Total samples selected Ns= 500

Airplanes that arrived on time Na = 482.

Airplanes that arrived late Nl = 500 - 482 = 18

Calculating the probability of an airplane arriving late:

P(L) = Nl/Ns

P(L) = 18/500

P(L) = 0.036

An event is deemed significant if its probability is equal to or less than 0.05.

As P(L) < 0.05

P(L) = Significant at 0.036

Answer:

4

Step-by-step explanation:

6-4+2*5=12

12/3

4

Answer:

The chance of completing the entire package installation in under 12 minutes is 0.1271.

Step-by-step explanation:

We define X as a normal distribution representing the time taken in seconds to install the software. According to the Central Limit Theorem, X is approximately normal, where the mean is 15 and variance is 15, giving a standard deviation of √15 = 3.873.

To find the probability of the total installation lasting less than 12 minutes, which equals 720 seconds, each installation should average under 720/68 = 10.5882 seconds. Thus, we seek the probability that X is less than 10.5882. To do this, we will apply W, the standard deviation value of X, calculated via the formula provided.

Utilizing , we reference the cumulative distribution function of the standard normal variable W, with values found in the attached file.

Given the symmetry of the standard normal distribution density function, we ascertain .

Consequently, the probability that the installation process for the entire package is completed within 12 minutes is 0.1271.

Answer:

Robyn's model is logical, while Mark's is illogical.

Step-by-step explanation:

This question doesn't require calculations. What we need to do is analyze each model logically.

Mark's

Mark's representation indicates 20 instead of 2, which signifies that 200 is ten times greater than 20, making it nonsensical.

Robyn's

Robyn's representation displays 2, suggesting that 200 is 100 times greater than 2, which is not only accurate but also reasonable since 100 * 2 equals 200.

Answer:

There is a probability of 24.51% that the weight of a bag exceeds the maximum permitted weight of 50 pounds.

Step-by-step explanation:

Problems dealing with normally distributed samples can be addressed using the z-score formula.

For a set with the mean and a standard deviation , the z-score for a measure X is calculated by

Once the Z-score is determined, we consult the z-score table to find the related p-value for this score. The p-value signifies the likelihood that the measured value is less than X. Since all probabilities total 1, calculating 1 minus the p-value gives us the probability that the measure exceeds X.

For this case

Imagine the weights of passenger bags are normally distributed with a mean of 47.88 pounds and a standard deviation of 3.09 pounds, thus

What probability exists that a bag’s weight will surpass the maximum allowable of 50 pounds?

That translates to

Thus

has a p-value of 0.7549.

<pthis indicates="" that="" src="https://tex.z-dn.net/?f=P%28X%20%5Cleq%2050%29%20%3D%200.7549" id="TexFormula10" title="P(X \leq 50) = 0.7549" alt="P(X \leq 50) = 0.7549" align="absmiddle" class="latex-formula">.

Additionally, we have that

There is a probability of 24.51% that the weight of a bag will exceed the maximum allowable weight of 50 pounds.

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