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muminat
3 years ago
5

How do i do solve this math problem?

Mathematics
2 answers:
erica [24]3 years ago
5 0
-3*5*7=-105
When a square root of an a expression is multiples by itself the result is that expression.
mestny [16]3 years ago
4 0

When you multiply two square roots together, you basically multiply the numbers within and put a squre root over that.

For example, in this case:

sqrt (5) * sqrt (5) = sqrt (25) = 5

Now that we have 5, we just multiply that to -3 and 7:

5 * -3 * 7 = -105

The answer is -105.

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Step-by-step explanation:

How much carpet is asking the area and area= LW

so Length(4) x Width(6) = 24

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Answer:

\frac{110}{3}    or     36\frac{2}{3}

Step-by-step explanation:

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3 years ago
Etermine the x-intercept and y intercept of the following equations.
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4 years ago
A small lawnmower company produced 1,500 lawnmowers in 2008. In an effort to determine how maintenance-free these units were, th
ikadub [295]

Answer:

The 95% confidence interval for the average number of years until the first major repair is (3.1, 3.5).

Step-by-step explanation:

The (1 - <em>α</em>)% confidence interval for the average using the finite correction factor is:

CI=\bar x\pm z_{\alpha/2}\cdot\frac{\sigma}{\sqrt{n}}\cdot\sqrt{\frac{N-n}{N-1}}

The information provided is:

N=1500\\n=183\\\sigma=1.47\\\bar x=3.3

The critical value of <em>z</em> for 95% confidence level is,

<em>z</em> = 1.96

Compute the 95% confidence interval for the average number of years until the first major repair as follows:

CI=\bar x\pm z_{\alpha/2}\cdot\frac{\sigma}{\sqrt{n}}\cdot\sqrt{\frac{N-n}{N-1}}

     =3.3\pm 1.96\times\frac{1.47}{\sqrt{183}}\times\sqrt{\frac{1500-183}{1500-1}}\\\\=3.3\pm 0.19964\\\\=(3.10036, 3.49964)\\\\\approx (3.1, 3.5)

Thus, the 95% confidence interval for the average number of years until the first major repair is (3.1, 3.5).

7 0
3 years ago
Fill in the blank with a constant, so that the resulting expression can be factored as the product of two linear expressions: 2a
Vitek1552 [10]

Answer:

2ab - 6a + 5b - 15

Step-by-step explanation:

Given

2ab - 6a + 5b + \_

Required

Fill in the gap to produce the product of linear expressions

2ab - 6a + 5b + \_

Split to 2

(2ab - 6a) + (5b + \_)

Factorize the first bracket

2a(b - 3) + (5b + \_)

Represent the _ with X

2a(b - 3) + (5b + X)

Factorize the second bracket

2a(b - 3) + 5(b + \frac{X}{5})

To result in a linear expression, then the following condition must be satisfied;

b - 3 = b + \frac{X}{5}

Subtract b from both sides

b - b- 3 = b - b+ \frac{X}{5}

- 3 = \frac{X}{5}

Multiply both sides by 5

- 3 * 5 = \frac{X}{5} * 5

X = -15

Substitute -15 for X in 2a(b - 3) + 5(b + \frac{X}{5})

2a(b - 3) + 5(b + \frac{-15}{5})

2a(b - 3) + 5(b - \frac{15}{5})

2a(b - 3) + 5(b - 3)

(2a + 5)(b - 3)

The two linear expressions are (2a+ 5) and (b - 3)

Their product will result in 2ab - 6a + 5b - 15

<em>Hence, the constant is -15</em>

3 0
3 years ago
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