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katrin2010 [14]

3 years ago

5

Can someone please help me solve number 13?

1 answer:

oksian1 [2.3K]3 years ago

6 0

If an exponent is negative you need to switch it to the other side.

So it would be 54v¼•w⅔

__________

18v⅔•¹¹/³•x5/7

Then you need to simplify it. So you would do 54÷18 to get 3.

NOTE: when you are simplifying it's dividing, so the answer you get goes where the larger number is. So in this case the 3 is where the 54 was.

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A=6 cm,b=9cm,and c =3 cm

Crazy boy [7]

The lengths of sides a and c add to 9 cm, the length of side b.
In a triangle, the the sum of the lengths of any two sides must be greater than the the length of the third side. These dimensions do not make a triangle.

Is this really what you meant? Why did you use capital A for a side? Or is A an angle, and it measures 6 deg, not 6 cm?

4 0

3 years ago

Y= [4x]<br> Constant Function<br> Quadratic Function<br> Cubic Function<br> Step Function

abruzzese [7]

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A. The set of happy people.

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3 years ago

Solve for x in the equation x2 - 12x+ 36 = 90

12345 [234]

Answer:

6 + 3 $\sqrt{10\\}$ or 6 - 3 $\sqrt{10\\}$

Step-by-step explanation:

x^2 - 12x + 36 = 90

x^2 - 12x - 54 = 0

x = $\frac{-b +/- \sqrt{b^{2} + 4ac }}{2a}$

a = 1, b= -12, c= -54

6 + 3 $\sqrt{10\\}$ or 6 - 3 $\sqrt{10\\}$

6 0

3 years ago

Math Help? 18 Points

alexgriva [62]

1. 43,300

2. 568,000

3. 910,000

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4 0

3 years ago

Read 2 more answers

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