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What are the x- and y- intercepts of the graph of 9x-3y = 18?

sineoko [7]

Step-by-step explanation:

9x - 3y = 18

Solve for y:

1. 3y= 9x + 18

2. y= 3x + 6 (This is the equation you will graph)

To graph:

Make a table for y = 3x +6.

If x =0, y=6

If x =1, y=9

If x =-1, y=3

You could get as many points as you like, however you need at least TWO, to be able to draw the line.

3 0

2 years ago

There is a set of 100 observations with a mean of 50 and a standard deviation of 0. What is the value of the smallest observatio

neonofarm [45]

In order to obtain a standard deviation of 0, all members of the data set should be equal.
This means that each of the 100 members in the data set is equal to 50 (the mean value).

Answer:
The smallest observation in the set is 50.

8 0

3 years ago

Read 2 more answers

1000 students participated in a survey on a university campus about their TV watching habits. All

nasty-shy [4]

Hi, you've asked an unclear question. However, I inferred you may want to know the actual number of students represented by the percentages of 27%, and 61%.

Explanation:

Finding percentage usually involves performing two operations; multiplication and division.

First, all (100%) of respondents said they watched TV at least at some point during the day.

Next, 27% of respondents stated that they only watched television during prime time hours, in which the actual number of students represented by the percentage is calculated by dividing 27 by 100 and multiplying by 1000 =

$\frac{27}{100} * 1000 = 270$ .

Finally, we are told 61% of respondents stated that they spend prime time hours in their dorm rooms. The actual number of students represented by the percentage is calculated by dividing 61 by 100 and multiplying by 1000 =

$\frac{61}{100} * 1000 = 610$

5 0

3 years ago

The tensile strength of stainless steel produced by a plant has been stable for a long time with a mean of 72 kg/mm2 and a stand

Elanso [62]

Answer:

95% confidence interval for the mean of tensile strength after the machine was adjusted is [73.68 kg/mm2 , 74.88 kg/mm2].

Yes, this data suggest that the tensile strength was changed after the adjustment.

Step-by-step explanation:

We are given that the tensile strength of stainless steel produced by a plant has been stable for a long time with a mean of 72 kg/mm 2 and a standard deviation of 2.15.

A machine was recently adjusted and a sample of 50 items were taken to determine if the mean tensile strength has changed. The mean of this sample is 74.28. Assume that the standard deviation did not change because of the adjustment to the machine.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean strength of 50 items = 74.28

$\sigma$ = population standard deviation = 2.15

n = sample of items = 50

$\mu$ = population mean tensile strength after machine was adjusted

Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level of

significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $74.28-1.96 \times {\frac{2.15}{\sqrt{50} } }$ , $74.28+1.96 \times {\frac{2.15}{\sqrt{50} } }$ ]

= [73.68 kg/mm2 , 74.88 kg/mm2]

Therefore, 95% confidence interval for the mean of tensile strength after the machine was adjusted is [73.68 kg/mm2 , 74.88 kg/mm2].

Yes, this data suggest that the tensile strength was changed after the adjustment as earlier the mean tensile strength was 72 kg/mm2 and now the mean strength lies between 73.68 kg/mm2 and 74.88 kg/mm2 after adjustment.

8 0

3 years ago

Given that u is the midpoint of on and v is the midpoint of rn select all statements that are true

Ira Lisetskai [31]

Answer:

6,9,15

It's a triple stack

8 0

3 years ago