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oksano4ka [1.4K]

3 years ago

13

A salesperson is paid $60 per week plus $4.50 per sale. THIS WEEK, the salesperson wants to earn more than $240. How many sales

must the salesperson make inorder to meet that goal?
Translated Inequality:

Inequality Solution:

Will the salesperson meet his/her goal with 40 sales? (answer yes or no):

1 answer:

inn [45]3 years ago

6 0

Answer:

42 sales plz mark as brainliest

Step-by-step explanation:

60 + 4.5(x) =250

60 + 4.5x =250

4.5x = 250-60

4.5x = 190

x = 190/4.5

x= 42 sales

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Determine the slope of the line that has the following coordinates: (5, 9)(11, - 3)

AlladinOne [14]

Answer:

$x_1 = 5, x_2 =11, y_1 =9, y_2 = -3$

The slope can be founded with this formula:

$m =\frac{y_2 -y_1}{x_2 -x_1}$

And replacing we got:

$m =\frac{-3 -9}{11-5}= -2$

And the best answer for this case would be :

$m=-2$

Step-by-step explanation:

For this case we have the following two points given:

$x_1 = 5, x_2 =11, y_1 =9, y_2 = -3$

The slope can be founded with this formula:

$m =\frac{y_2 -y_1}{x_2 -x_1}$

And replacing we got:

$m =\frac{-3 -9}{11-5}= -2$

And the best answer for this case would be :

$m=-2$

7 0

3 years ago

masha68 [24]

Answer:

51.9 cm²

Step-by-step explanation:

From the diagram attached,

Area of the white region(segment)(A) = Area minor sector- area of the triangle

A = (πr²∅/360°)-(1/2r²sin∅)............... Equation 1

Where r = radius of the circle, Ф = reflex angle formed at the center of the circle, π = pie

From the question,

Given: r = 7 cm, Ф = 150°

Constant: π = 22/7

Substitute these values into equation 1

A = [(22/7)×7²×150/360]-[(1/2)×7²×sin150]

A = 64.17-12.25

A = 51.92

A = 51.9 cm²

5 0

2 years ago

Find f(x) and g(x) so that the function can be described as y = f(g(x)).<br><br> y = 4/x^2+9

dlinn [17]

I'm assuming all of (x^2+9) is in the denominator. If that assumption is correct, then,

One possible answer is $f(x) = \frac{4}{x}, \ \ g(x) = x^2+9$

Another possible answer is $f(x) = \frac{4}{x+9}, \ \ g(x) = x^2$

There are many ways to do this. The idea is that when we have f( g(x) ), we basically replace every x in f(x) with g(x)

So in the first example above, we would have

$f(x) = \frac{4}{x}\\\\f( g(x) ) = \frac{4}{g(x)}\\\\f( g(x) ) = \frac{4}{x^2+9}$

In that third step, g(x) was replaced with x^2+9 since g(x) = x^2+9.

Similar steps will happen with the second example as well (when g(x) = x^2)

4 0

3 years ago

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marishachu [46]

Answer: it’s B

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

An article contained the following observations on degree of polymerization for paper specimens for which viscosity times concen

devlian [24]

Answer:

(a) A 95% confidence interval for the population mean is [433.36 , 448.64].

(b) A 95% upper confidence bound for the population mean is 448.64.

Step-by-step explanation:

We are given that article contained the following observations on degrees of polymerization for paper specimens for which viscosity times concentration fell in a certain middle range:

420, 425, 427, 427, 432, 433, 434, 437, 439, 446, 447, 448, 453, 454, 465, 469.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean = $\frac{\sum X}{n}$ = 441

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 14.34

n = sample size = 16

$\mu$ = population mean

<em>Here for constructing a 95% confidence interval we have used One-sample t-test statistics as we don't know about population standard deviation.</em>

<em />

<u>So, 95% confidence interval for the population mean, </u> $\mu$ <u> is ;</u>

P(-2.131 < $t_1_5$ < 2.131) = 0.95 {As the critical value of t at 15 degrees of

freedom are -2.131 & 2.131 with P = 2.5%}

P(-2.131 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.131) = 0.95

P( $-2.131 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.131 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.131 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.131 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

<u>95% confidence interval for</u> $\mu$ = [ $\bar X -2.131 \times {\frac{s}{\sqrt{n} } }$ , $\bar X +2.131 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $441-2.131 \times {\frac{14.34}{\sqrt{16} } }$ , $441+2.131 \times {\frac{14.34}{\sqrt{16} } }$ ]

= [433.36 , 448.64]

(a) Therefore, a 95% confidence interval for the population mean is [433.36 , 448.64].

The interpretation of the above interval is that we are 95% confident that the population mean will lie between 433.36 and 448.64.

(b) A 95% upper confidence bound for the population mean is 448.64 which means that we are 95% confident that the population mean will not be more than 448.64.

6 0

3 years ago