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2 years ago

Solve for x for #5 and 6

Mathematics

1 answer:

Alex777 [14]2 years ago

7 0

Answer:

In 5:

UM and MS are equal since the diagonals US and VT have perpendicularly bisected each other.

So,15=6x-3

15+3=6x

18=6x

3=x

In 6:

ET=TC

8=x-4

8+4=x

12=x

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A new lawn has a perimeter of 278ft. And a width of 64ft. A bag of grass seed will cover 384ft of ground. How many bags of grass

Anit [1.1K]

Answer:

13 bags

Step-by-step explanation:

Area of the lawn: A=lw

w= 64 ft

P= 2(l+w)= 278 ft ⇒

l= 278/2- 64= 75 ft

A= 64*75= 4800 ft²

bags of seed required:

4800/384= 12.5 so 13 full bags needed

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3 years ago

Evaluate f(x)=5•3 ( to the power of x) for x=4

GenaCL600 [577]

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3 years ago

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A cylinder and a cone have the same volume. The cylinder has radius x and height y. The cone has radius 3x. Find the height of t

timurjin [86]

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3 years ago

The scores of students on the ACT college entrance exam in a recent year had the normal distribution with mean  =18.6 and stand

Maurinko [17]

Answer:

a) 33% probability that a single student randomly chosen from all those taking the test scores 21 or higher.

b) 0.39% probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher

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 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 $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 18.6, \sigma = 5.9$

a) What is the probability that a single student randomly chosen from all those taking the test scores 21 or higher?

This is 1 subtracted by the pvalue of Z when X = 21. So

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

$Z = \frac{21 - 18.6}{5.4}$

$Z = 0.44$

$Z = 0.44$ has a pvalue of 0.67

1 - 0.67 = 0.33

33% probability that a single student randomly chosen from all those taking the test scores 21 or higher.

b) The average score of the 76 students at Northside High who took the test was x =20.4. What is the probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher?

Now we have $n = 76, s = \frac{5.9}{\sqrt{76}} = 0.6768$