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marin [14]
2 years ago
7

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 ⇒

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  • A= 64*75= 4800 ft²

bags of seed required:

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3 years ago
Evaluate f(x)=5•3 ( to the power of x) for x=4
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The answer is 50625 because 5*3 is 15 and take that to the power of 4
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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
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Vcyl = Vcone
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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

This probability is 1 subtracted by the pvalue of Z when X = 20.4. So

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

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{20.4 - 18.6}{0.6768}

Z = 2.66

Z = 2.66 has a pvalue of 0.9961

1 - 0.9961 = 0.0039

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

4 0
2 years ago
Please show your work and check!!
ad-work [718]

Step 1. Add 6 to both sides

a ≥ 1 + 6

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2 years ago
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