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2 years ago

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2. Ms. Rich is creating a walkway around her rectangular garden. If the walkway will be 3 feet wide, and the garden is 10ft by 1

2ft, what will the total area of the walkway be? Hint: Draw a picture!
can anybody help me

1 answer:

Yuki888 [10]2 years ago

4 0

The area of the walkway is 168 square feet if Ms. Rich is creating a walkway around her rectangular garden. If the walkway will be 3 feet wide, and the garden is 10ft by 12ft.

<h3>What is the area of the rectangle?</h3>

It is defined as the space occupied by the rectangle, which is planner 2-dimensional geometry.

The formula for finding the area of a rectangle is given by:

Area of rectangle = length × width

The dimensions of the garden = 10 ft by 12 ft

Area = 10×12 = 120 square feet

The width of the walkway = 3 feet

If we create a walkway around the garden, it will be a rectangle around the garden.

So the dimensions for the bigger rectangle = (3+10+3) ft by (3+12+3) ft

Area of the bigger rectangle = 16×18 = 288 square feet

So the area of the walkway = 288 - 120 = 168 square feet

Thus, the area of the walkway is 168 square feet if Ms. Rich is creating a walkway around her rectangular garden. If the walkway will be 3 feet wide, and the garden is 10ft by 12ft.

Learn more about the rectangle here:

brainly.com/question/15019502

#SPJ1

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Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Part a

Let X the random variable that represent the amount of time that people spend at Grover Hot Springs of a population, and for this case we know the distribution for X is given by:

$X \sim N(63,18)$

Where $\mu=63$ and $\sigma=18$

Part b

We are interested on this probability

$P(X>90)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

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

If we apply this formula to our probability we got this:

$P(X>90)=P(\frac{X-\mu}{\sigma}>\frac{90-\mu}{\sigma})=P(Z>\frac{90-63}{18})=P(Z>1.5)$

And we can find this probability using the complement rule and with the normal standard table or excel:

$P(Z>1.5)=1-P(Z$

Part c

We are interested on this probability

$P(X$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

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

If we apply this formula to our probability we got this:

$P(X$

And we can find this probability using the normal standard table or excel:

$P(Z$

Part d

If we apply this formula to our probability we got this:

$P(60$

And we can find this probability with this difference:

$P(-0.167$

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.

$P(-0.167$

Part e

Q1

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.75$ (a)

$P(X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.25 of the area on the left and 0.75 of the area on the right it's z=-0.674. On this case P(Z<-0.674)=0.25 and P(z>-0.674)=0.75

If we use condition (b) from previous we have this:

$P(X$

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=-0.674$

And if we solve for a we got

$a=63 -0.674*18=50.87$

So the value of height that separates the bottom 25% of data from the top 75% is 50.87.

Q3

Since the distribution is symmetrical we repaeat the procedure for Q1 but now with z= 0.674

$z=0.674$

And if we solve for a we got

$a=63 +0.674*18=75.13$

And then the IQR = 75.13-50.87=24.26

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