Monica is a lawyer who earns $8,322.65 every month. She spends $830.98 on rent, $210.36 on food, and $120.45 on gas every month.

Answer:

The proportion of goats at fair Y exceeds that of goats at fair X.

Step-by-step explanation:

From the bar chart,  we see there were 100 goats at both fairs X and Y.

Calculating the percentage of goats at each fair we find:

Fair X = 100/645 * 100 = 15.5%

Fair Y = 100/590 * 100 = 16.9%

This indicates that fair Y had a higher percentage of goats compared to fair X.

Response: Using the Pythagorean theorem, we can set up the equation as 14^2=9^2+x^2

                                     -9^2    -9^2

To address this issue, we will utilize the formula for determining the distance from a point to a line.

The formula is:

distance = | a x + b y + c | / sqrt (a^2 + b^2)

We have the line equation:

y = 2 x + 4

Rearranging it results in:

<span>y – 2 x – 4 = 0 --> a = -2, b = 1, c = -4</span>

The coordinates given are:

(-4, 11) = (x, y)

Substituting into the distance formula:

distance = | -2 * -4 + 1 * 11 + -4 | / sqrt [(- 2)^2 + (1)^2]

distance = 15 / sqrt (5)

distance ≈ 6.7

<span>Thus, the tree is approximately 6.7 ft from the zip line.</span>

Answer:

Step-by-step explanation:

Proceed to your table

Response:

Detailed explanation:

Greetings!

You have the variable

X: Area eligible for painting with a can of spray paint (feet²)

This variable is normally distributed with a mean of μ= 25 feet² and a standard deviation of δ= 3 feet²

As this variable has a normal distribution, it needs to be converted into the standard normal form to utilize tabulated cumulative probabilities.

a.

P(X>27)

The first step involves standardizing the X value using Z= (X-μ)/ δ ~N(0;1)

P(Z>(27-25)/3)

P(Z>0.67)

Having determined the Z value, you can find it in the table, but since the table includes probabilities for , the following conversion must be applied:

P(Z>0.67)= 1 - P(Z≤0.67)= 1 - 0.74857= 0.25143

b.

A sample of 20 cans was taken, and you need to ascertain the probability of averaging a coverage area of 540 feet².

The sample mean maintains the same distribution as its source variable, but its variance is influenced by sample size, thus it is normally distributed with parameters:

X[bar]~N(μ;δ²/n)

To cover 540 feet² with 20 cans, the average coverage must be approximately 540/20= 27 feet² per can.

c.

No, if the distribution is not normal and skewed, the normal distribution should not be applied for calculating probabilities. While the central limit theorem might approximate the sampling distribution to normal when the sample size is 30 or larger, that isn’t applicable here.

I trust this information is helpful!