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2 years ago

What is y-intercept of the graph

Mathematics

1 answer:

Kipish [7]2 years ago

7 0

Answer:

1a. y-intercept: 12

1b. slope: -3/2

1c. equation: y = -3/2x +12

2a. y-intercept: -9

2b. slope: 2

2c. equation: y = 2x -9

Step-by-step explanation:

A) We observe the pattern to be <em>x-values in the table increase by 2, while y-values in the table decrease by 3</em>. We notice the first x-value is 2, so extending the table upward to x=0 would tell us the y-intercept. That is, adding 3 to the first y-value will give the y-intercept as (x, y) = (0, 12).

B) We have already observed that the "rise" (change in y) is -3 for each "run" (change in x) of 2. The slope is the ratio of these changes:

slope = m = rise/run = -3/2

C) From the above, we know that m=-3/2 and b=12. Putting these values into the equation for the line gives ...

y = -3/2x +12

A) We observe the pattern to be <em>y-values increase by 2 while x-values increase by 1</em>. As before, we can find the point that would go before the first one shown in the table. It will have an x-value of 0 and a y-value of -9.

the y-intercept is -9

the slope is 2/1 = 2

the equation is y = 2x -9

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A 98% confidence interval for the mean assembly time is [21.34, 26.49] .

Step-by-step explanation:

We are given that a sample of 40 times yielded an average time of 23.92 minutes, with a sample standard deviation of 6.72 minutes.

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 average time = 23.92 minutes

s = sample standard deviation = 6.72 minutes

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

<u>So, a 98% confidence interval for the population mean, </u> $\mu$ <u> is; </u>

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P(-2.426 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.426) = 0.98

P( $-2.426 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.426 \times {\frac{s}{\sqrt{n} } }$ ) = 0.98

P( $\bar X-2.426 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.426 \times {\frac{s}{\sqrt{n} } }$ ) = 0.98

<u>98% confidence interval for</u> $\mu$ = [ $\bar X-2.426 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.426 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $23.92-2.426 \times {\frac{6.72}{\sqrt{40} } }$ , $23.92+2.426 \times {\frac{6.72}{\sqrt{40} } }$ ]

= [21.34, 26.49]

Therefore, a 98% confidence interval for the mean assembly time is [21.34, 26.49] .

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