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shepuryov [24]
3 years ago
11

lateisha sold a total of 136,000 of office equipment last month. if her commission is 7.8% of sales, how much commission did she

earn.
Mathematics
1 answer:
Irina18 [472]3 years ago
7 0

Answer:

$136,000*7.8%= $10,608

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Thickness measurements of ancient prehistoric Native American pot shards discovered in a Hopi village are approximately normally
rusak2 [61]

Answer:

(a) The probability that the thickness is less than 3.0 mm is 0.119.

(b) The probability that the thickness is more than 7.0 mm is 0.119.

(c) The probability that the thickness is between 3.0 mm and 7.0 mm is 0.762.

Step-by-step explanation:

We are given that thickness measurements of ancient prehistoric Native American pot shards discovered in a Hopi village are approximately normally distributed, with a mean of 5.0 millimeters (mm) and a standard deviation of 1.7 mm.

Let X = <u><em>thickness measurements of ancient prehistoric Native American pot shards discovered in a Hopi village.</em></u>

So, X ~ Normal(\mu=5.0,\sigma^{2} =1.7^{2})

The z-score probability distribution for the normal distribution is given by;

                           Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean thickness = 5.0 mm

           \sigma = standard deviation = 1.7 mm

(a) The probability that the thickness is less than 3.0 mm is given by = P(X < 3.0 mm)

    P(X < 3.0 mm) = P( \frac{X-\mu}{\sigma} < \frac{3.0-5.0}{1.7} ) = P(Z < -1.18) = 1 - P(Z \leq 1.18)

                                                           = 1 - 0.8810 = <u>0.119</u>

The above probability is calculated by looking at the value of x = 1.18 in the z table which has an area of 0.881.

(b) The probability that the thickness is more than 7.0 mm is given by = P(X > 7.0 mm)

    P(X > 7.0 mm) = P( \frac{X-\mu}{\sigma} > \frac{7.0-5.0}{1.7} ) = P(Z > 1.18) = 1 - P(Z \leq 1.18)

                                                           = 1 - 0.8810 = <u>0.119</u>

The above probability is calculated by looking at the value of x = 1.18 in the z table which has an area of 0.881.

(c) The probability that the thickness is between 3.0 mm and 7.0 mm is given by = P(3.0 mm < X < 7.0 mm) = P(X < 7.0 mm) - P(X \leq 3.0 mm)

    P(X < 7.0 mm) = P( \frac{X-\mu}{\sigma} < \frac{7.0-5.0}{1.7} ) = P(Z < 1.18) = 0.881

    P(X \leq 3.0 mm) = P( \frac{X-\mu}{\sigma} \leq \frac{3.0-5.0}{1.7} ) = P(Z \leq -1.18) = 1 - P(Z < 1.18)

                                                           = 1 - 0.8810 = 0.119

The above probability is calculated by looking at the value of x = 1.18 in the z table which has an area of 0.881.

Therefore, P(3.0 mm < X < 7.0 mm) = 0.881 - 0.119 = 0.762.

4 0
3 years ago
How do I do this?? 7-11
mihalych1998 [28]

Answer: slope intercept is y=mx+b form s b= the y intercept and x will just be x and the m is usually a fraction which tells you the slope so what you need to do is mark these points on a line graph draw a line from one to the other and every where it crosses over a point is where you can find the slope and the y intercept giving you the answer



7 0
3 years ago
Is the point (2,8) a solution of the equation y = 13x + 12?<br> O Yes<br> O No
vovikov84 [41]

Answer:

O No

Explanation:

Given equation: y = 13x + 12

To check if (2, 8) is a solution of the given equation.

Substitute x and y value in equation and check if it is true.

\rightarrow y = 13x+12

(x, y) = (2, 8)

\rightarrow 8 = 13(2)+12

simplify

\rightarrow 8 = 38

This following statement is false and hence (2, 8) is not a solution.

6 0
1 year ago
Name the sets of numbers to which -5 belongs to
kipiarov [429]
It would belong to integers, rational, and real numbers.
5 0
3 years ago
Determine the equations of the vertical and horizontal asymptotes, if any, for y=x^3/(x-2)^4
djverab [1.8K]

Answer:

Option a)

Step-by-step explanation:

To get the vertical asymptotes of the function f(x) you must find the limit when x tends k of f(x). If this limit tends to infinity then x = k is a vertical asymptote of the function.

\lim_{x\to\\2}\frac{x^3}{(x-2)^4} \\\\\\lim_{x\to\\2}\frac{2^3}{(2-2)^4}\\\\\lim_{x\to\\2}\frac{2^3}{(0)^4} = \infty

Then. x = 2 it's a vertical asintota.

To obtain the horizontal asymptote of the function take the following limit:

\lim_{x \to \infty}\frac{x^3}{(x-2)^4}

if \lim_{x \to \infty}\frac{x^3}{(x-2)^4} = b then y = b is horizontal asymptote

Then:

\lim_{x \to \infty}\frac{x^3}{(x-2)^4} \\\\\\lim_{x \to \infty}\frac{1}{(\infty)} = 0

Therefore y = 0 is a horizontal asymptote of f(x).

Then the correct answer is the option a) x = 2, y = 0

3 0
3 years ago
Read 2 more answers
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