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Anastaziya [24]
3 years ago
11

15Points!! Factor completely. 12x² +28x

Mathematics
1 answer:
Zigmanuir [339]3 years ago
4 0

Answer:

4x(3x+7)

Step-by-step explanation:

12x^2+28x\\12xx+28x\\\\4\cdot \:3xx+4\cdot \:7x\\4x\left(3x+7\right)

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11. Identify the properties below.<br> a. 5 + 6 = 6 + 5<br> b. 7 + (3 + 9) = (7 + 9) + 3
jenyasd209 [6]
A is commutative property
B is associative property
Explanation:
5 0
3 years ago
three roommates Ava, Ben, and Cora went grocery shopping online. Ava spent three times as much Ben and half as much Cora. If tog
Mamont248 [21]

Answer:

Cora's spend will be = $ 16.66

Step-by-step explanation:

Let y be the amount Ben spent

As Ava spent three times Ben will be 3y and half as much Cora(2y)

so the equation becomes

y + 2y + 3y = 50

6y = 50

y = 50/6

y = 8.33

So Cora's spend will be : 2y = 2(8.33) = 16.66 units currency

Note: if currency is in $

Then Cora's spend will be = $ 16.66

8 0
3 years ago
Please please help me What is 5% of 200?
Leni [432]

Answer:

10

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
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guapka [62]

Answer:

Step-by-step explanation:

abs ( -2(5-(-1)) / 3 )

= abs ( -2(5+1) / 3 )

= abs ( -2(6) /3)

= abs (-12 /3)

= 4

(3(-1) -- 1) /2

= (-3 -1) /2

= -4 / 2

= -2

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5 0
3 years ago
Question Help Suppose that the lifetimes of light bulbs are approximately normally​ distributed, with a mean of 5656 hours and a
koban [17]

Answer:

a)3.438% of the light bulbs will last more than 6262 hours.

b)11.31% of the light bulbs will last 5252 hours or less.

c) 23.655% of the light bulbs are going to last between 5858 and 6262 hours.

d) 0.12% of the light bulbs will last 4646 hours or less.

Step-by-step explanation:

Normally distributed problems can be solved by the z-score formula:

On a normaly distributed set with mean \mu and standard deviation \sigma, the z-score of a value X is given by:

Z = \frac{X - \mu}{\sigma}

After we find the value of Z, we look into the z-score table and find the equivalent p-value of this score. This is the probability that a score will be LOWER than the value of X.

In this problem, we have that:

The lifetimes of light bulbs are approximately normally​ distributed, with a mean of 5656 hours and a standard deviation of 333.3 hours.

So \mu = 5656, \sigma = 333.3

(a) What proportion of light bulbs will last more than 6262 ​hours?

The pvalue of the z-score of X = 6262 is the proportion of light bulbs that will last less than 6262. Subtracting 100% by this value, we find the proportion of light bulbs that will last more than 6262 hours.

Z = \frac{X - \mu}{\sigma}

Z = \frac{6262 - 5656}{333.3}

Z = 1.82

Z = 1.81 has a pvalue of .96562. This means that 96.562% of the light bulbs are going to last less than 6262 hours. So

P = 100% - 96.562% = 3.438% of the light bulbs will last more than 6262 hours.

​(b) What proportion of light bulbs will last 5252 hours or​ less?

This is the pvalue of the zscore of X = 5252

Z = \frac{X - \mu}{\sigma}

Z = \frac{5252- 5656}{333.3}

Z = -1.21

Z = -1.21 has a pvalue of .1131. This means that 11.31% of the light bulbs will last 5252 hours or less.

(c) What proportion of light bulbs will last between 5858 and 6262 ​hours?

This is the pvalue of the zscore of X = 6262 subtracted by the pvalue of the zscore X = 5858

For X = 6262, we have that Z = 1.81 with a pvalue of .96562.

For X = 5858

Z = \frac{X - \mu}{\sigma}

Z = \frac{5858- 5656}{333.3}

Z = 0.61

Z = 0.61 has a pvalue of .72907.

So, the proportion of light bulbs that will last between 5858 and 6262 hours is

P = .96562 - .72907 = .23655

23.655% of the light bulbs are going to last between 5858 and 6262 hours.

​(d) What is the probability that a randomly selected light bulb lasts less than 4646 ​hours?

This is the pvalue of the zscore of X = 4646

Z = \frac{X - \mu}{\sigma}

Z = \frac{4646- 5656}{333.3}

Z = -3.03

Z = -3.03 has a pvalue of .0012. This means that 0.12% of the light bulbs will last 4646 hours or less.

5 0
3 years ago
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