Help on math problem

Mathematics

1 answer:

Anvisha [2.4K]3 years ago

5 0

Das is similarity

the 2 triangles are similar because of math ( don't remember which theorem or whatever)

so since similar

we have to find the bottom length
use right triangle properties
a²+b²=c²
14²+5.5²=c²
196+30.25=c²
226.25=c²
√226.25=c

ten, we do similarity

the hyptonuse of smaller triangle to base of smaller triangle is √226.25:5.5
since the triangles are similar, that should be equal to the ratio of the hypotonuse of the largger triangle to the base, which is x:√226.25

so
x/√226.25=√226.25/5.5
times both sides by √226.25
x=226.25/5.5
x≈41.1363636

the height of screen is 6 less than total
41.13636-6=35.13636

height of screen is about 35ft

You might be interested in

g The average midterm score of students in a certain course is 70 points. From the past experience it is known that the midterm

spayn [35]

Answer:

0.98 = 98% probability that the average midterm score of these students is at most 75 points.

Step-by-step explanation:

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

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, 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 normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .