There were 5,317 previously owned homes sold in a western city in the year 2000. The distribution of the sales prices of these h

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.

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.

Answer:

A Type II error occurs when the null hypothesis is not rejected, even when the alternative hypothesis is actually valid.

The implication is that the new method may be dismissed or altered despite it being a real enhancement.

Step-by-step explanation:

Answer:

B

Step-by-step explanation:

If two quantities have a direct variation, their graph will originate from the point of intersection at the origin.

Among all graphs, only Nikiya's intersects the origin, thus the answer is B.

Answer:

Tan(b°) = 3/4, which is equivalent to three fourths (C).

Step-by-step explanation:

Triangle JKL is a right triangle with angle K being the right angle and angle L equal to b°.

We will employ SOHCAHTOA principles from trigonometry to calculate the sides' values.

For triangle ∆JKL:

Sin(b°) = opposite/hypotenuse

Sin(b°) = 3/5

Cos(b°) = adjacent/hypotenuse

Cos(b°) = 4/5

Tan(b°) = 3/4.

From the earlier information, we have the values for each side of triangle ∆JKL.

Please refer to the attached diagram for the triangle (1).

Triangle ∆JKL is scaled with a factor of 2.

This implies multiplying each side of ∆JKL by 2, resulting in:

Opposite side = 2(3) = 6

Adjacent side = 2(4) = 8

Hypotenuse = 2(5) = 10.

To find tan(b°) for the new triangle, we use the tangent ratio:

Tan(b°) = opposite/adjacent.

Tan(b°) = 6/8.

Tan(b°) = 3/4.

Find the diagram for the new triangle included (2).

Diagram 3 provides a depiction of both triangles.

As shown, the angle remains unchanged when a triangle is scaled by a factor.

Thus, we conclude that tan(b°) = three fourths (C).