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ki77a [65]
3 years ago
5

You have 6,000 with which to build a rectangular enclosure with fencing

Mathematics
1 answer:
Dominik [7]3 years ago
7 0
What is the question asking
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What is the figure sir/ma’am
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Math Help? 14 Points
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The Solution is 5.68 x 10^5
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3 years ago
Find the equation of the line using the point-slope formula. Write all the final equations
vivado [14]

Answer:

y=\frac{1}{2}x+\frac{5}{2}

Step-by-step explanation:

We are going to find the slope by lining up the points vertically and subtract, then put 2nd difference over first.

Also you could just use \frac{y_2-y_1}{x_2-x_1}.  It is the same thing.

(  5  ,   5)

-(  1  ,   3)

--------------

  4      2

So the slope is 2/4 or 1/2 after reducing.

The point-slope form a line is

y-y_1=m(x-x_1) with a point on the line (x_1,y_1)=(1,3) given and with slope m=\frac{1}{2} given.

y-3=\frac{1}{2}(x-1)

We are going to solve this for y and simplify what we can because our goal is y=mx+b; this is slope-intercept form.  It is called that because it tells us the slope,m, and the y-intercept,b.

Distribute:

y-3=\frac{1}{2}x-\frac{1}{2}

Add 3 on both sides:

y=\frac{1}{2}x-\frac{1}{2}+3

Simplify:

y=\frac{1}{2}x+\frac{5}{2}

4 0
3 years ago
Can you guys please help ( for did the letter half cut out is Y
BabaBlast [244]

Answer:

\overline{OQ}

OQ

​

. Given OQ=2x,OQ=2x, OP=x,OP=x, and PQ=10,PQ=10, determine \overline{OQ}.

Oq

8 0
2 years ago
The weight of turkeys is normally distributed with a mean of 22 pounds and a standard deviation of 5 pounds. a. Find the probabi
Masja [62]

Answer:

a. 0.443

b. 0.023

Step-by-step explanation:

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.

The weight of turkeys is normally distributed with a mean of 22 pounds and a standard deviation of 5 pounds.

This means that \mu = 22, \sigma = 5

a. Find the probability that a randomly selected turkey weighs between 20 and 26 pounds.

This is the pvalue of Z when X = 26 subtracted by the pvalue of Z when X = 20. So

X = 26

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

Z = \frac{26 - 22}{5}

Z = 0.8

Z = 0.8 has a pvalue of 0.788

X = 20

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

Z = \frac{20 - 22}{5}

Z = -0.4

Z = -0.4 has a pvalue of 0.345

0.788 - 0.345 = 0.443

The answer is 0.443

b. Find the probability that a randomly selected turkey weighs below 12 pounds.

This is the pvalue of Z when X = 12. So

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

Z = \frac{12 - 22}{5}

Z = -2

Z = -2 has a pvalue of 0.023

The answer is 0.023.

8 0
3 years ago
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