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Aleks [24]
3 years ago
7

Kamila has a rectangular lawn that measures 80 by 150 feet. She wants to lay gravel made from colored rock in a border, 6 feet w

ide, all around the lawn. Create a diagram that shows Kamila’s rectangular lawn and the gravel border. Assign variables to any unknown sides and label the diagram. Use your diagram to determine how many square feet of the gravel border will surround the lawn. Kamila decided that the depth of the gravel needs to be 2 inches. What is the total volume of gravel needed for the border? (Hint: dimensions must be in terms of the same unit before you can calculate the volume) D. A large bag of colored rock contains of a cubic yard and costs $30. What is the total cost of the rocks needed for the border? (Hint: the volume of the border must be in terms of a cubic yard before you can calculate the cost)
Mathematics
1 answer:
beks73 [17]3 years ago
7 0
A) Diagram

Below is a sketch of the recctangular lawn (the inside rectangle) and the gravel border.


  6 ft + 150 yards + 6 ft
-------------------------------
|                                    |
|     ---------------------      |
|     |                        |     |
|     |                        |     |   6 ft + 80 yard + 6 ft
|     |                        |     |
|      ---------------------     |
|                                    |
|------------------------------|

B) How many square feet of the gravel will surround the lawn?

Conversion: 1 yard = 3 feet

=> 80 yard = 80*3 feet = 240 feet

=> 150 yard = 150 * 3 feet = 450 feet

Area of the external rectangle = (6 ft + 240 ft + 6 ft) (6 ft + 450 ft + 6 ft) = 252 ft * 462 ft = 116,424 ft^2

Area of the internal rectangle = 240 ft * 450 ft = 108,000 ft^2

Area of the gravel border = 116,424 ft^2 - 108,000 ft^2 = 8424 ft^2

answer: 8424 ft^2

C) volume

volume = area of the base * depth

2 inches = 2/12 ft = 1/6 ft

=> volume = 8424 ft^2 * 1/6 ft = 1404 ft^3

answer: 1404 ft^3

C) cost

1 yard^3 = $ 30

1yard ^3 = (3 feet)^3 = 27 ft^3

=> cost = $ 30 / 27 ft^3

=> 1404 ft^3 * $30 / 27 ft^3 = $ 1560

answer: $ 1560
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Pls answer asap it's urgent
Korolek [52]
Interior angle add up to 360, so you add
(a+10) + (3a+10) + (5a-10) + (3a-10) = 12a

12a =360
a= 30
30+10=40
3(30)+10=100
5(30)-10=140
3(30)-10=80

You can double check they're correct by doing 40+100+140+80, which equals 360.
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#X and y inetercept

If a equation given y=mx+b

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A coffee packaging plant claims that the mean weight of coffee in its containers is at least 32 ounces. A random sample of 15 co
Luda [366]

Answer:

We conclude that the mean weight of coffee in its containers is at least 32 ounces which means that the data support the claim.

Step-by-step explanation:

We are given that a coffee packaging plant claims that the mean weight of coffee in its containers is at least 32 ounces.

A random sample of 15 containers were weighed and the mean weight was 31.8 ounces with a sample standard deviation of 0.48 ounces.

Let \mu = <u><em>mean weight of coffee in its containers.</em></u>

SO, Null Hypothesis, H_0 : \mu \geq 32 ounces     {means that the mean weight of coffee in its containers is at least 32 ounces}

Alternate Hypothesis, H_A : \mu < 32 ounces      {means that the mean weight of coffee in its containers is less than 32 ounces}

The test statistics that would be used here <u>One-sample t-test statistics</u> as we don't know about population standard deviation;

                             T.S. =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean weight = 31.8 ounces

             s = sample standard deviation = 0.48 ounces

             n = sample of containers = 15

So, <u><em>the test statistics</em></u>  =  \frac{31.8 -32}{\frac{0.48}{\sqrt{15} } }  ~ t_1_4  

                                      =  -1.614

The value of t test statistics is -1.614.

<u>Now, at 0.01 significance level the t table gives critical value of -2.624 at 14 degree of freedom for left-tailed test.</u>

Since our test statistic is more than the critical value of t as -1.614 > -2.624, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u><em>we fail to reject our null hypothesis</em></u>.

Therefore, we conclude that the mean weight of coffee in its containers is at least 32 ounces which means that the data support the claim.

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3 years ago
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