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REY [17]
3 years ago
7

I need help pleas and thnk u

Mathematics
1 answer:
Mazyrski [523]3 years ago
7 0
Smallest x intercept is (-8,0)
Biggest x-intercept is (7,0)
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Which equation best represents the line graphed above? x = –1 x = 1 y = –1 y = x – 1
White raven [17]

I think you forgot the graph. Please add it.

8 0
3 years ago
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The daily revenue at a university snack bar has been recorded for the past five years. Records indicate that the mean daily reve
Scorpion4ik [409]

Answer:

sample standard deviation is $50

Step-by-step explanation:

Given a  population standard deviation of $500 and a random sample of size, n=100.

-The sample standard deviation is calculated using the formula:

s=\sigma/\sqrt{n}

Where:

  • s is the sample standard deviation.
  • n is the sample size.
  • \sigma is the population standard deviation

Therefore:

s=\sigma/\sqrt{n}\\\\=500\sqrt{100}\\\\=50

Hence, the sampling distribution has a mean of $1,500 and a standard deviation of $50

8 0
3 years ago
What the slope intercept form of 5x-2y=6
grin007 [14]

5x-2y=6

Slope intercept form y = mx + b

So

5x-2y=6

2y = 5x - 6

 y = 5/2 x - 3

Answer

Slope intercept form:  y = 5/2 x - 3

4 0
4 years ago
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Graphing round each value
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3 years ago
Estimate the total percentage of complaints about parking difficulties and the location of the store. (nearest whole percent)
AysviL [449]

Answer:

Therefore the total  Complains about parking difficulties=32.26%

Therefore the total  Complains about the location of the store=25.81%

Step-by-step explanation:

Given ,

Complains about difficulty parking = 250

Complains about isolated location = 200

Complains about ruled sales clerks=150

Complains about too expensive=125

Complains about confusing layout= 50

Total complains = (250+200+150+125+50)

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Therefore the total  Complains about parking difficulties

=\frac{250}{775}\times 100 %

=32.26%

Therefore the total  Complains about the location of the store

=\frac{200}{775}\times 100 %

=25.81%

4 0
3 years ago
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