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2 years ago

11

Given the linear equation 2x + y = 6, write another linear equation in two variables such that these two equations when represen

ted geometrically form parallel lines.

1 answer:

Furkat [3]2 years ago

5 0

Answer:

The family of lines that are parallel to $2\cdot x + y = 6$ is of the form $2\cdot x + y = k, \,k\neq 6$ . A possible solution is $2\cdot x + y = -2$ .

Step-by-step explanation:

Let be $2\cdot x + y = 6$ the equation of a line, another line is parallel to it if and only if it is of the form:

$2\cdot x + y = k, \,k\neq 6$ (1)

Then, a possible solution is $2\cdot x + y = -2$ .

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3 years ago

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Part I - To help consumers assess the risks they are taking, the Food and Drug Administration (FDA) publishes the amount of nico

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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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3 years ago

In a standard deck of 52 playing cards, what is the probability

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Step-by-step explanation:

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What is the total weight of 10 buses that each weigh 12 1/3 tons?

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Answer:

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Step-by-step explanation:

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3 years ago

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