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3 years ago

Which of the following points does not lie on the line y = -2x +5?

Mathematics

2 answers:

saw5 [17]3 years ago

8 0

Answer:

C (3, 1 )

Step-by-step explanation:

To determine if a point lies on the line, substitute the x- coordinate of the point into the equation and if the value obtained equals the y- coordinate of the point then it lies on the line, otherwise not on line.

A (4 , - 3)

y = - 2(4) + 5 = - 8 + 5 = - 3 ← lies on line

B (- 2, 9 )

y = - 2(- 2) + 5 = 4 + 5 = 9 ← lies on line

C (3, 1 )

y = - 2(3) + 5 = - 6 + 5 = - 1 ≠ 1 ← not on line

D (- 1, 7 )

y = - 2(- 1) + 5 = 2 + 5 = 7 ← lies on line

E (2, 1 ) = - 2(2) + 5 = - 4 + 5 = 1 ← lies on line

Thus (3, 1 ) does not lie on the line → C

igor_vitrenko [27]3 years ago

6 0

Answer:

C and E are not on the line.

Step-by-step explanation:

Just substitute for the x and the y and see which ordered pair fits and which does not:

If Left side of the equation = right side then the point fits.

A. y = -3

-2x+5 = -2(4) + 5 = -3 So A is on the line

B. y = 9

-2(-2) + 5 = 9 So this point does lie on the line.

C. y = 1

-2(3) + 5 = -1 No.

D. y = 7

-1(-1) + 5 = 7 YES.

E. y = 1

-2(1) + 5 = 3 NO.

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A new catalyst is being investigated for use in the production of a plastic chemical. Ten batches of the chemical are produced.

zzz [600]

Answer:

$CI=(66.54,78.46)$

Step-by-step explanation:

Given : A new catalyst is being investigated for use in the production of a plastic chemical. Ten batches of the chemical are produced. The mean yield of the 10 batches is 72.5% and the standard deviation is 5.8%. Assume the yields are independent and approximately normally distributed.

To find : A 99% confidence interval for the mean yield when the new catalyst is used ?

Solution :

Let X be the yield of the batches.

We have given, n=10 , $\bar{X}=72.5\%$ , s=5.8%

Since the size of the sample is small.

We will use the student's t statistic to construct a 995 confidence interval.

$\bar X\pm t_{n-1,\frac{\alpha}{2}}\frac{s}{\sqrt n}$

From the t-table with 9 degree of freedom for $\frac{\alpha}{2}=0.005$

$t_{n-1,\frac{\alpha}{2}}=t_{9,0.005}$

$t_{n-1,\frac{\alpha}{2}}=3.250$

The 99% confidence interval is given by,

$CI=72.5 \pm 3.25\frac{5.8}{\sqrt{10}}$

$CI=72.5 \pm 5.96$

$CI=(72.5+5.96),(72.5-5.96)$

$CI=(66.54,78.46)$

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Answer:

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Step-by-step explanation:

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Let's put it into our equation

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