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Angelina_Jolie [31]
3 years ago
11

Sam is planning to fence his back yard. Every side of the yard except for the side along the house is to be fenced, and fencing

costs $3/yd and can only be bought in whole yards. (Note that m ZNPM-28, and the side of his yard opposite the house measures 35 yd) How much will the fencing cost? Explain how you found your answer.
Mathematics
1 answer:
Morgarella [4.7K]3 years ago
4 0

Answer:

The cost of the fencing is $273

Step-by-step explanation:

* Lets explain how to solve the problem

- Sam is planning to fence his back yard

- Every side of the yard except for the side along the house is to

 be fenced

- The fencing costs $3/yd and can only be bought in whole yards

- The side of his yard opposite the house measures 35 yd

- The other two sides of his yard are 28 yd

* Lets solve the problem

∵ The length of the yard is the sum of the sides whose to be fenced

∴ The length of the yard = 28 + 35 + 28 = 91 yd

∵ The cost of the fencing is $3 per yard

∴ The cost = 91 × 3 = 273

∴ The cost of the fencing is $273

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15 + 8(7)= x
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3 years ago
A cell phone plan cost $200 to start. Then there is a $50 charge each month.
andreev551 [17]

Answer:

1) $250

2) $200 + $50x

Step-by-step explanation:

1)

$200 (start up fee) + $50 * ( 1 month ) = $200 + $50 = $250

2)

$200 (start up fee) + $50 * ( x months ) = $200 + $50x

Hope this helps :)

5 0
3 years ago
I’m confused please helppp
Alborosie
For the first one we see that the shape is that of a parabola and is translated down a few units, We can eliminate linear, exponential
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I hope that helps
4 0
3 years ago
A Simple random sample of 100 8th graders at a large suburban middle school indicated that 84% of them are involved with some ty
Setler [38]

Answer: a) (0.755, 0.925)

Step-by-step explanation:

Let p be the population proportion of 8th graders are involved with some type of after school activity.

As per given , we have

n= 100

sample proportion: \hat{p}=0.84

Significance level : \alpha= 1-0.98=0.02

Critical z-value : z_{\alpha/2}=2.33  (using z-value table)

Then, the 98% confidence interval that estimates the proportion of them that are involved in an after school activity will be :-

\hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}

i.e. 0.84\pm (2.33)\sqrt{\dfrac{0.84(1-0.84)}{100}}

i.e. \approx0.84\pm 0.085

i.e. (0.84- 0.085,\ 0.84+ 0.085)=(0.755,\ 0.925)

Hence, the 98% confidence interval that estimates the proportion of them that are involved in an after school activity : a) (0.755, 0.925)

3 0
3 years ago
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That would be angle 2. It is on the opposite of angle 7 and it is inside the two parallel lines

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5 0
3 years ago
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