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pentagon [3]
3 years ago
5

Given the equation: 3x^2-12=0

Mathematics
1 answer:
cluponka [151]3 years ago
7 0

Step-by-step explanation:

3x²-12=0

3(x²-4)=0

3(x²-2²)=0

3(x+2)(x-2)=0

hope it helps.

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2. Darnell bought(7.7 kilograms of blueberries at
Ugo [173]

Answer:4.75961

Step-by-step explanation:

7.7x0.35=2.695x1.79=<u>4.75961</u>

3 0
2 years ago
A firm producing 7 units of output has an average total cost of Rs. 150 and has to pay Rs.350 to its fixed factors of production
maria [59]

Rs 100 of the average total cost is made up of variable costs.

Step-by-step explanation:

Given:

Number of output the firm produces= 7 units

Average cost of the output= Rs. 150

fixed factors of production = Rs.350

To Find:

How much of the average total cost is made up of variable costs=?

Solution:

we know that,

Average total cost= total cost/ number of output units produced

substituting the values, we get

150=\frac{\text{Total cost}}{7}

Total cost= 1050

we know that Total fixed cost = 350

Total cost = Total fixed cost + Total variable cost

plug in the known values.

1050= 350 + Total variable cost

Total variable cost = 1050-350

Total variable cost =700

For 7th unit \frac{700}{7} = 100

5 0
3 years ago
Need help on this ASAPPPPP
lord [1]

Answer: B

Step-by-step explanation:

Over to the right 5 and down 2

3 0
3 years ago
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What is 1 inch to 6 feet
docker41 [41]

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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

5 0
2 years ago
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