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3 years ago

9

A garden is a rectangle 100 feet long and 50 feet wide. The portion of the garden devoted to herbs is 10 feet by 10 feet. The re

st of the area is for vegetables. How many square feet are given for growing vegetables?

2 answers:

postnew [5]3 years ago

4 0

Answer:

Area for vegetable = 4900 square ft

Step-by-step explanation:

The garden is a rectangle . According to the question the length of the garden is 100 ft and the width is 50 ft. The portion of area devoted to herb is 10 ft by 10 ft. That means the area devoted to herbs is a square. The area devoted for vegetable can be computed when one subtract area for herbs from the area of the garden.

Area of the rectangle garden = length × width

length = 100 ft

width = 50 ft

Area of the rectangle garden = 100 × 50

Area of the rectangle garden = 5000 square ft

Area devoted for herbs (square) = length²

length = 10 ft

Area devoted for herbs (square) = 10²

Area devoted for herbs (square) = 10 × 10

Area devoted for herbs (square) = 100 square ft

Area for vegetable = Area of garden - Area for herbs

Area for vegetable = 5000 - 100

Area for vegetable = 4900 square ft

lys-0071 [83]3 years ago

3 0

4,900 sq feet are given for growing vegetables

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3 years ago

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Answer:

a. The positive difference between Nelson's height and the population mean is: $\\ \lvert 68-70.7 \rvert = \lvert 70.7-68 \rvert\;in = 2.7\;in$ .

b. The difference found in part (a) is 1.174 standard deviations from the mean (without taking into account if the height is above or below the mean).

c. Nelson's z-score: $\\ z = -1.1739 \approx -1.174$ (Nelson's height is <em>below</em> the population's mean 1.174 standard deviations units).

d. Nelson's height is <em>usual</em> since $\\ -2 < -1.174 < 2$ .

Step-by-step explanation:

The key concept to answer this question is the z-score. A <em>z-score</em> "tells us" the distance from the population's mean of a raw score in <em>standard deviation</em> units. A <em>positive value</em> for a z-score indicates that the raw score is <em>above</em> the population mean, whereas a <em>negative value</em> tells us that the raw score is <em>below</em> the population mean. The formula to obtain this <em>z-score</em> is as follows:

$\\ z = \frac{x - \mu}{\sigma}$ [1]

Where

$\\ z$ is the <em>z-score</em>.

$\\ \mu$ is the <em>population mean</em>.

$\\ \sigma$ is the <em>population standard deviation</em>.

From the question, we have that:

Nelson's height is 68 in. In this case, the raw score is 68 in $\\ x = 68$ in.
$\\ \mu = 70.7$ in.
$\\ \sigma = 2.3$ in.

With all this information, we are ready to answer the next questions:

a. What is the positive difference between Nelson's height and the mean?

The positive difference between Nelson's height and the population mean is (taking the absolute value for this difference):

$\\ \lvert 68-70.7 \rvert = \lvert 70.7-68 \rvert\;in = 2.7\;in$ .

That is, <em>the positive difference is 2.7 in</em>.

b. How many standard deviations is that [the difference found in part (a)]?

To find how many <em>standard deviations</em> is that, we need to divide that difference by the <em>population standard deviation</em>. That is:

$\\ \frac{2.7\;in}{2.3\;in} \approx 1.1739 \approx 1.174$

In words, the difference found in part (a) is 1.174 <em>standard deviations</em> from the mean. Notice that we are not taking into account here if the raw score, <em>x,</em> is <em>below</em> or <em>above</em> the mean.

c. Convert Nelson's height to a z score.

Using formula [1], we have

$\\ z = \frac{x - \mu}{\sigma}$

$\\ z = \frac{68\;in - 70.7\;in}{2.3\;in}$

$\\ z = \frac{-2.7\;in}{2.3\;in}$

$\\ z = -1.1739 \approx -1.174$

This z-score "tells us" that Nelson's height is <em>1.174 standard deviations</em> <em>below</em> the population mean (notice the negative symbol in the above result), i.e., Nelson's height is <em>below</em> the mean for heights in the club presidents of the past century 1.174 standard deviations units.

d. If we consider "usual" heights to be those that convert to z scores between minus2 and 2, is Nelson's height usual or unusual?

Carefully looking at Nelson's height, we notice that it is between those z-scores, because:

$\\ -2 < z_{Nelson} < 2$

$\\ -2 < -1.174 < 2$

Then, Nelson's height is <em>usual</em> according to that statement.

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solniwko [45]

Given:

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Which type of triangle is ΔWXY by its sides.

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$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Using distance formula, we get

$WX=\sqrt{(-3-(-10))^2+(-1-4)^2}$

$WX=\sqrt{(-3+10)^2+(-5)^2}$

$WX=\sqrt{(7)^2+(-5)^2}$

$WX=\sqrt49+25}$

$WX=\sqrt{74}$

Similarly,

$XY=\sqrt{\left(-5-\left(-3\right)\right)^2+\left(11-\left(-1\right)\right)^2}=2\sqrt{37}$

$WY=\sqrt{\left(-5-\left(-10\right)\right)^2+\left(11-4\right)^2}=\sqrt{74}$

Now,

$WX=WY$

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3 years ago