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bogdanovich [222]
3 years ago
12

a home improvement store advertised 60 sq ft of flooring for $287 including an $80 installation fee write an equation to represe

nt the cost of one square foot of flooring
Mathematics
1 answer:
Leno4ka [110]3 years ago
6 0
60c+80=287

Let c stand for the cost of one square foot. with 60 square feet you get 60c. then add 80 for the installation fee.
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Simplify this question m=12
amm1812
12x12=144
3x144=432
The answer is 432.
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2 years ago
Find the value of C: <br>6x + Cy = 18<br>3x + 2y = 8
Natalija [7]
I think C is 6, I’m not sure if thats right though. But if X=2 and Y=1
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3 years ago
............................................whats 10000000000000000000x1000000000000000000000000
koban [17]

Answer:

10,000,000,000,000,000,000,000,000,000,000,000,000,000,000

Step-by-step explanation:

10,000,000,000,000,000,000 x 1,000,000,000,000,000,000,000,000 =

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3 0
2 years ago
Read 2 more answers
The number of texts per day by students in a class is normally distributed with a 
kobusy [5.1K]

Answer:

1, 2, 6

Step-by-step explanation:

The z score shows by how many standard deviations the raw score is above or below the mean. The z score is given by:

z=\frac{x-\mu}{\sigma} \\\\where\ x=raw\ score,\mu=mean, \ \sigma=standard\ deviation

Given that mean (μ) = 130 texts, standard deviation (σ) = 20 texts

1) For x < 90:

z=\frac{x-\mu}{\sigma} \\\\z=\frac{90-130}{20} =-2

From the normal distribution table, P(x < 90) = P(z < -2) = 0.0228 = 2.28%

Option 1 is correct

2) For x > 130:

z=\frac{x-\mu}{\sigma} \\\\z=\frac{130-130}{20} =0

From the normal distribution table, P(x > 130) = P(z > 0) = 1 - P(z < 0) = 1 - 0.5 = 50%

Option 2 is correct

3) For x > 190:

z=\frac{x-\mu}{\sigma} \\\\z=\frac{190-130}{20} =3

From the normal distribution table, P(x > 3) = P(z > 3) = 1 - P(z < 3) = 1 - 0.9987 = 0.0013 = 0.13%

Option 3 is incorrect

4)  For x < 130:

z=\frac{x-\mu}{\sigma} \\\\z=\frac{130-130}{20} =0

For x > 100:

z=\frac{x-\mu}{\sigma} \\\\z=\frac{100-130}{20} =-1.5

From the normal table, P(100 < x < 130) = P(-1.5 < z < 0) = P(z < 0) - P(z < 1.5) = 0.5 - 0.0668 = 0.9332 = 93.32%

Option 4 is incorrect

5)  For x = 130:

z=\frac{x-\mu}{\sigma} \\\\z=\frac{130-130}{20} =0

Option 5 is incorrect

6)  For x = 130:

z=\frac{x-\mu}{\sigma} \\\\z=\frac{160-130}{20} =1.5

Since 1.5 is between 1 and 2, option 6 is correct

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2 years ago
What is solution to 3/5=z/10
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Z=6

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3 years ago
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