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Usimov [2.4K]
3 years ago
13

What two numbers multiply -14 and add up to 13

Mathematics
1 answer:
Galina-37 [17]3 years ago
8 0

Answer:

14*-1

Step-by-step explanation:

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Answer:

log (8)

Step-by-step explanation:

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4 years ago
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Solve: 2a + 5 = 23<br> :((
artcher [175]

Answer: 9

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5 0
3 years ago
A farmer wants to fence in a rectangular plot of land adjacent to the north wall of his barn. No fencing is needed along the bar
Zina [86]

Answer:

The dimensions for the plot that would enclose the most area are as follows:

Width  = 116.67 ft

Length = 175 ft.

Step-by-step explanation:

Note: Kindly refer to the attached diagram where:

x = Length of the rectangular area

y = Width of the rectangular area.

The fencing along the west side of the plot is shared with a neighbor who will split the cost of that portion of the fence. This implies that it is $10 per foot. Otherwise, it will be $20 per foot. Since no fencing is needed along the barn, it implies that there is no fence along the northern length.

Since the farmer is not willing to spend more than $7000, we have:

10y + 20x + 20y = 7000

20x + 30y = 7000

Divide through by 20, we have:

x + 1.5y = 350

x = 350 - 1.5y …………………………. (1)

The fenced area is

A = x * y ………………………………….. (2)

Substituting equation (1) into (2), we have:

A = (350 - 1.5y)y

A = 350y - 1.5y^2 ……………. (3)

To maximize A, equation (3) is differentiated with respect to y and equated it to 0 as follows:

That is,

A'(y) = 350 - 3y = 0

Solving for y, we have:

y = 350/3

y = 116.67 ft

Substituting y = 116.67 for y in equation (1), we have:

x = 350 - (1.5 * 116.67)

x = 175ft

Therefore, the dimensions for the plot that would enclose the most area are as follows:

Width  = 116.67 ft

Length = 175 ft.

3 0
3 years ago
The lifetimes of a certain type of light bulbs follow a normal distribution. If approximately 2.5% of the bulbs have lives excee
spin [16.1K]

Answer:

Mean = 339 hours

Standard deviation = 54 hours

Step-by-step explanation:

We are given that approximately 2.5% of the bulbs have lives exceeding 445 hours

So,P(x>445) = 0.025

P(x<445) =1- 0.025 =0.975

 z value for 0.975 is 1.96

Formula : z=\frac{x-\mu}{\sigma}

So, 1.96=\frac{445-\mu}{\sigma}

1.96 \sigma =445-\mu--A

Now we are given that  approximately 16% have lives exceeding 393 hours,

So,P(x>393) = 0.16

P(x<393) =1- 0.16 =0.84

 z value for 0.84 is 1

Formula : z=\frac{x-\mu}{\sigma}

So, 1=\frac{393-\mu}{\sigma}

1 \sigma =393-\mu ---B

Solve A and B

Substitute the value of \sigmafrom B in A

1.96 (393 -\mu) =445-\mu

770.28 -1.96\mu=445-\mu

770.28 -445=1.96\mu-\mu

325.28=0.96\mu

\frac{325.28}{0.96}=\mu

338.83=\mu

Substitute the value in B

\sigma =393-338.83

\sigma =54.17

So, mean = 339 hours

Standard deviation = 54 hours

5 0
3 years ago
Find the mean absolute deviation of the set of data. 5, 10, 12, 15, 20
vladimir2022 [97]
The answer to your question is 4.08
6 0
3 years ago
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