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Akimi4 [234]
3 years ago
11

Please Help !! Show All Work !!

Mathematics
1 answer:
GrogVix [38]3 years ago
4 0

Answer:

1) (4,4)  2) (4,0)   3) (6,1)

Step-by-step explanation:

(\frac{5+13}{2} )(\frac{7+1}{2} )\\(\frac{8}{2} )(\frac{8}{2} )\\(4,4)

(\frac{-3+7}{2} )(\frac{-3+3}{2} )\\(\frac{4}{2} )(\frac{0}{2} )\\(4,0)

\frac{-5+x_2}{2} =1 multiply 1 by 2=2 then write(+5)-5+x_2=1(+5)\\x_2 = 6

\frac{3+y_2}{2} =2 multiply 2 by 2= 4 then write (-3)3+y_2=4(-3)\\y_2=1

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34(5f−3)=38 How do i find the value of f?
Oduvanchick [21]

f=14/17 or 0.82352941

8 0
3 years ago
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An English professor assigns letter grades on a test according to the following scheme. A: Top 9% of scores B: Scores below the
sergeinik [125]

Answer:

The minimum score required for an A grade is 83.

Step-by-step explanation:

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.

Mean of 72.3 and a standard deviation of 8.

This means that \mu = 72.3, \sigma = 8

Find the minimum score required for an A grade.

This is the 100 - 9 = 91th percentile, which is X when Z has a pvalue of 0.91, so X when Z = 1.34.

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

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X - 72.3 = 1.34*8

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The minimum score required for an A grade is 83.

5 0
3 years ago
A rope of length 18 feet is arranged in the shape of a sector of a circle with central angle O radians, as shown in the
creativ13 [48]

Answer:

A(\theta)=\frac{162 \theta}{(\theta+2)^2}

Step-by-step explanation:

The picture of the question in the attached figure

step 1

Let

r ---> the radius of the sector

s ---> the arc length of sector

Find the radius r

we know that

2r+s=18

s=r \theta

2r+r \theta=18

solve for r

r=\frac{18}{2+\theta}

step 2

Find the value of s

s=r \theta

substitute the value of r

s=\frac{18}{2+\theta}\theta

step 3

we know that

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A=\pi r^{2}

The complete circle subtends a central angle of 2π radians

so

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Let

A ---> the area of sector with central angle theta

\frac{\pi r^{2} }{2\pi}=\frac{A}{\theta} \\\\A=\frac{r^2\theta}{2}

substitute the value of r

A=\frac{(\frac{18}{2+\theta})^2\theta}{2}

A=\frac{162 \theta}{(\theta+2)^2}

Convert to function notation

A(\theta)=\frac{162 \theta}{(\theta+2)^2}

6 0
4 years ago
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Similarly, for a d-heap, the height is ceiling(log_d(n)).



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VLD [36.1K]
The answer is 24 dollars.
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