Angle C shows the angle to be formed by the ramp from the ground, which is 15 degrees. Also, from the ground, it’s going to be four feet tall, which is line AB. The top of the ramp is point A, which makes line AC the entire length of the ramp. Since we have a reference angle (angle C) and two sides, the opposite and the hypotenuse, we shall apply the trigonometric ratio.

SinC = opposite/hypotenuse

Sin 15 = 4/b

By cross multiplication we now have

b = 4/Sin15

b = 4/0.2588

b = 15.4599

Approximately b = 15.5

Therefore the length of the ramp would be 15.5 feet

6 0

3 years ago

Which linear inequality will not have a shared solution set with the graphed linear inequality?

valentina_108 [34]

Answer:

C. y > 5/3x + 2

7 0

3 years ago

Read 2 more answers

How do i rearrange the formula <br><br> w = x^2 - 2yz<br><br> to make y the subject?

iogann1982 [59]

Solve for y by simplifying both sides of the equation, then isolating the variable.
y = x^2/2^z - w/2z

7 0

3 years ago

A population has mean 187 and standard deviation 32. If a random sample of 64 observations is selected at random from this popul

Zina [86]

Answer:

11.51% probability that the sample average will be less than 182

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

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

$Z = \frac{X - \mu}{\sigma}$

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.

Central Limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$