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s344n2d4d5 [400]
3 years ago
6

A family built a house and wanted to put down in their rectangular yard. The sod costs $0.32 per square foot plus 10 per square

yard to lay the sod. The front yard is 30 by 50 feet. How much will it cost to purchase and lay the sod?
Mathematics
1 answer:
sammy [17]3 years ago
3 0

Answer:

The total cost of purchasing and laying sods in the rectangular yard is $2,146.67  

Step-by-step explanation:

We are given the following in the question:

Dimensions of rectangular yard:

Length = 50 feet

Width = 30 feet

Area of rectangular yard =

A = \text{Length}\times \text{Width}\\A = 50\times 30\\A  = 1500\text{ square feet}

Cost of sod = $0.32 per square foot

Total cost of sod =

C_1 = \text{Unit cost of sod}\times \text{Area of yard}\\C_1 = 0.32\times 1500\\C_1 = 480\$

Cost of laying sods = $10 per square yard

Total cost of laying sod =

C_2 = \text{Unit cost of laying sod}\times \text{Area of yard}\\\\\text{1 square yard}=\dfrac{1}{9}\text{Square foot}\\\\C_2 = 10\times \dfrac{1500}{9}\\\\C_2 = 1666.67\$

Total cost of purchasing and laying the sod =

C = C_1 + C_2\\C = 480 +1666.67\\C = \$2,146.67

Thus, the total cost of purchasing and laying sods in the rectangular yard is $2,146.67

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Step-by-step explanation:

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3 years ago
How do I delete this question? I did not mean to ask it.
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Step-by-step explanation:

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A horse's mane grows at the rate of 6/5 centimeters per month. How many months will it take the horse's mane to grow 24/10 centi
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24/10
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3 years ago
(1 - 5q ) +2 (2.5q +8 )
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Consider a value to be significantly low if its z score less than or equal to minus−2 or consider a value to be significantly hi
ziro4ka [17]

Answer:

If the weight is higher than 5.8886 gr would be considered significantly high

If the weight is lower than 5.6121 gr would be considered significantly low

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".  

Solution to the problem

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

X \sim N(5.75241,0.06281)  

Where \mu=5.75241 and \sigma=0.06281

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}

For the case when z =-2 we can do this:

-2 = \frac{X-5.75241}{0.06281}

And if we solve for X we got:

X = 5.75241 -2*0.06281 =5.6121

And for the other case when Z=2 we have:

2 = \frac{X-5.75241}{0.06281}

And if we solve for X we got:

X = 5.75241 +2*0.06281 =5.8886

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