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Murljashka [212]
4 years ago
14

The vertices of are D(7, 3), E(4, -3), and F(10, -3). Write a proof to prove that is isosceles.

Mathematics
1 answer:
nataly862011 [7]4 years ago
6 0
<span>The vertices of are D(7, 3), E(4, -3), and F(10, -3)

Prove tat DE = DF
Length of a segment = </span>√[(x₂ - x₁) +(y₂ -y₁)]

DE = √([4-7)² + (-3 -3)²] = √45 = 3√5
DF = √[(10-7)² +(-3 -3)²] = √45 b= 3√5
Hence DE = DF = 3√5 and DEF is  ISOSCELES 
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IRINA_888 [86]

Answer:

you would have to add 3 to 1 cuz there is a one with the x and since 3 is negative that how you would get y=-3x+1

Step-by-step explanation:

8 0
3 years ago
What is the percent equivalent of 3/2 using a pattern
olganol [36]
1/2 is 50 percent. if 3/2 is 1/2 X 3, 50 X 3  is equal to 150. 3/2 is 150 percent
6 0
3 years ago
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Consider the question of whether the home team wins more than half of its games in the National Basketball Association. Suppose
spayn [35]

Answer:

0.0037 = 0.37% probability that the home team would win 65% or more of its games in a simple random sample of 80 games

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 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 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}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

The home team therefore wins 50% of its games

This means that p = 0.5

Determine the probability that the home team would win 65% or more of its games in a simple random sample of 80 games

Sample of 80 means that n = 80 and, by the Central Limit Theorem:

\mu = p = 0.65

s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.5*0.5}{80}} = 0.0559

This probability is 1 subtracted by the pvalue of Z when X = 0.65. So

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

By the Central Limit Theorem

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

Z = \frac{0.65 - 0.5}{0.0559}

Z = 2.68

Z = 2.68 has a pvalue of 0.9963

1 - 0.9963 = 0.0037

0.0037 = 0.37% probability that the home team would win 65% or more of its games in a simple random sample of 80 games

7 0
3 years ago
The marketing club at school is opening a student store. They randomly survey 50 students about how much money they spend on lun
Luden [163]

Answer:

The expected value for a student to spend on lunch each day = $5.18

Step-by-step explanation:

Given the data:

Number of students______$ spent

2 students______________$10

1 student________________$8

12 students______________$6

23 students______________$5

8 students_______________$4

4 students_______________$3

Sample size, n = 50.

Let's first find the value on each amount spent with the formula:

\frac{num. of students}{sample size} * dollar spent

Therefore,

For $10:

\frac{2}{50} * 10 = 0.4

For $8:

\frac{1}{50} * 8 = 0.16 l

For $6:

\frac{12}{50} * 6 = 1.44

For $5:

\frac{23}{50} * 5 = 2.3

For $4:

\frac{8}{50} * 4 = 0.64

For $3:

\frac{4}{50} * 3 = 0.24

To find the expected value a student spends on lunch each day, let's add all the values together.

Expected value =

$0.4 + $0.16 + 1.44 +$2.3 + $0.64 + $0.24

= $5.18

Therefore, the expected value for a student to spend on lunch each day is $5.18

7 0
3 years ago
Convert 8 7/9 yard into in
ZanzabumX [31]

Answer:

316 in.

Step-by-step explanation:

12 inches = 1 foot

8 feet. = 96 inches

And, add 7/9 of a yard.

Your answer is 316 in.

Hope this helps!

3 0
3 years ago
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