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olga55 [171]
3 years ago
5

A charity organization had to sale 18 tickets to their fundraiser Just to cover necessary production cost.They sold each ticket

for $45.Let y represent the net profit in dollars when they have sold x tickets
Mathematics
1 answer:
Harman [31]3 years ago
8 0

A charity organization had to sale 18 tickets to their fundraiser Just to cover necessary production cost.They sold each ticket for $45.Let y represent the net profit in dollars when they have sold x tickets

Which of the following information about the graph of the relationship is given?

Choose 1 answer:

A

Slope and x-intercept

B

Slope and y-intercept

C

Slope and a point that is not an intercept

D

x-intercept and y-intercept

E

y-intercept and a point that is not an intercept

F

Two points that are not intercepts

Answer:

A. Slope and x-intercept

Step-by-step explanation:

We are told that each Ticket cost $45 , so the number of ticket sold should be multiplied by the cost of 1 ticket , Also; we were informed that x tickets were sold, this implies that the amount will be $45 multiplied by x , which is $45x. To find the profit (y) , then we must find the cost of 18 tickets which will give us the cost price and then subtract it from the selling price ($45x).

NOW:

Given that :

a charity organization had to sell 18 tickets

and one ticket cost for $45

Then 18 tickets = $45 × 18 = $810

Let y represent the net profit in dollars when they have sold x tickets

So : y = 45 x - 810

Relating the above equation with equation of a straight line  y = mx +c ,

where :

m and c = the numbers on the gradient line   ; &

y = slope

x =  intercept where the graph crosses y - axis

Thus; the information given is the :Slope and x-intercept

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Answer:

The geometric mean of 2 and 8 comes out to equal 4.

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3 years ago
1. Find the value of x.<br> 55°<br> (4x - 5)°<br> O<br> 26<br> O
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Answer:

26

Step-by-step explanation:

The sum of angles in a triangle is 180 degrees, use this and set up an equation;

x + ( 4x -5 ) + 55 = 180

Simplify

x + ( 4x - 5 ) + 55 = 180

5x + 50 = 180

Inverse operations;

5x + 50 = 180

    -50     - 50

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2 years ago
Part I - To help consumers assess the risks they are taking, the Food and Drug Administration (FDA) publishes the amount of nico
IRINA_888 [86]

Answer:

(I) 99% confidence interval for the mean nicotine content of this brand of cigarette is [24.169 mg , 30.431 mg].

(II) No, since the value 28.4 does not fall in the 98% confidence interval.

Step-by-step explanation:

We are given that a new cigarette has recently been marketed.

The FDA tests on this cigarette gave a mean nicotine content of 27.3 milligrams and standard deviation of 2.8 milligrams for a sample of 9 cigarettes.

Firstly, the Pivotal quantity for 99% 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 nicotine content = 27.3 milligrams

            s = sample standard deviation = 2.8 milligrams

            n = sample of cigarettes = 9

            \mu = true mean nicotine content

<em>Here for constructing 99% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.</em>

<u>Part I</u> : So, 99% confidence interval for the population mean, \mu is ;

P(-3.355 < t_8 < 3.355) = 0.99  {As the critical value of t at 8 degree

                                      of freedom are -3.355 & 3.355 with P = 0.5%}  

P(-3.355 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 3.355) = 0.99

P( -3.355 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu} < 3.355 \times {\frac{s}{\sqrt{n} } } ) = 0.99

P( \bar X-3.355 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+3.355 \times {\frac{s}{\sqrt{n} } } ) = 0.99

<u />

<u>99% confidence interval for</u> \mu = [ \bar X-3.355 \times {\frac{s}{\sqrt{n} } } , \bar X+3.355 \times {\frac{s}{\sqrt{n} } } ]

                                          = [ 27.3-3.355 \times {\frac{2.8}{\sqrt{9} } } , 27.3+3.355 \times {\frac{2.8}{\sqrt{9} } } ]

                                          = [27.3 \pm 3.131]

                                          = [24.169 mg , 30.431 mg]

Therefore, 99% confidence interval for the mean nicotine content of this brand of cigarette is [24.169 mg , 30.431 mg].

<u>Part II</u> : We are given that the FDA tests on this cigarette gave a mean nicotine content of 24.9 milligrams and standard deviation of 2.6 milligrams for a sample of n = 9 cigarettes.

The FDA claims that the mean nicotine content exceeds 28.4 milligrams for this brand of cigarette, and their stated reliability is 98%.

The Pivotal quantity for 98% 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 nicotine content = 24.9 milligrams

            s = sample standard deviation = 2.6 milligrams

            n = sample of cigarettes = 9

            \mu = true mean nicotine content

<em>Here for constructing 98% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.</em>

So, 98% confidence interval for the population mean, \mu is ;

P(-2.896 < t_8 < 2.896) = 0.98  {As the critical value of t at 8 degree

                                       of freedom are -2.896 & 2.896 with P = 1%}  

P(-2.896 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 2.896) = 0.98

P( -2.896 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu} < 2.896 \times {\frac{s}{\sqrt{n} } } ) = 0.98

P( \bar X-2.896 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+2.896 \times {\frac{s}{\sqrt{n} } } ) = 0.98

<u />

<u>98% confidence interval for</u> \mu = [ \bar X-2.896 \times {\frac{s}{\sqrt{n} } } , \bar X+2.896 \times {\frac{s}{\sqrt{n} } } ]

                                          = [ 24.9-2.896 \times {\frac{2.6}{\sqrt{9} } } , 24.9+2.896 \times {\frac{2.6}{\sqrt{9} } } ]

                                          = [22.4 mg , 27.4 mg]

Therefore, 98% confidence interval for the mean nicotine content of this brand of cigarette is [22.4 mg , 27.4 mg].

No, we don't agree on the claim of FDA that the mean nicotine content exceeds 28.4 milligrams for this brand of cigarette because as we can see in the above confidence interval that the value 28.4 does not fall in the 98% confidence interval.

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