6 xy ( 1/2 x ^2 - 1/3 xy + 1/6 ^2 )

What is the value of the rational expression below when x is equal to 3 (17+x)\(x+2)

elena55 [62]

The value of the expression is 4.

We can find this by taking the original equation and plugging in 3 every time we see an x.

$\frac{17 + x}{x + 2}$

$\frac{17 + 3}{3 + 2}$

$\frac{20}{5}$

4

4 0

3 years ago

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Im not sure where to start with this one...

Nina [5.8K]

The volume of a cube is side length cubed (s^3)
So if you know the volume then take the cube root of 1/512... which is 1/8

So each side of the cube is 1/8.

There are 6 congruent sides on a cube (picture a “dice”). Do the area of each side would be (1/8)(1/8)= 1/64 then multiply that area by 6 and reduce ... you get 6/64 = 3/32 for the surface area

3 0

3 years ago

1-73.

cricket20 [7]

36inch - 2 feet

72inch - 4 feet

144inch - 8 feet

18 X 2 = 36

1 + 1 = 2

18 x 4 = 72

1 x 4 = 4

18 x 8 = 144

1 x 8 = 8

8 0

3 years ago

An article contained the following observations on degree of polymerization for paper specimens for which viscosity times concen

devlian [24]

Answer:

(a) A 95% confidence interval for the population mean is [433.36 , 448.64].

(b) A 95% upper confidence bound for the population mean is 448.64.

Step-by-step explanation:

We are given that article contained the following observations on degrees of polymerization for paper specimens for which viscosity times concentration fell in a certain middle range:

420, 425, 427, 427, 432, 433, 434, 437, 439, 446, 447, 448, 453, 454, 465, 469.

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 mean = $\frac{\sum X}{n}$ = 441

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 14.34

n = sample size = 16

$\mu$ = population mean

Here for constructing a 95% confidence interval we have used One-sample t-test statistics as we don't know about population standard deviation.

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

P(-2.131 < $t_1_5$ < 2.131) = 0.95 {As the critical value of t at 15 degrees of

freedom are -2.131 & 2.131 with P = 2.5%}

P(-2.131 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.131) = 0.95

P( $-2.131 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.131 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.131 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.131 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X -2.131 \times {\frac{s}{\sqrt{n} } }$ , $\bar X +2.131 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $441-2.131 \times {\frac{14.34}{\sqrt{16} } }$ , $441+2.131 \times {\frac{14.34}{\sqrt{16} } }$ ]

= [433.36 , 448.64]

(a) Therefore, a 95% confidence interval for the population mean is [433.36 , 448.64].

The interpretation of the above interval is that we are 95% confident that the population mean will lie between 433.36 and 448.64.

(b) A 95% upper confidence bound for the population mean is 448.64 which means that we are 95% confident that the population mean will not be more than 448.64.

6 0

3 years ago

Kate paid $27 for a dress that was on sale at 10% off. What was the original price for this dress?

wolverine [178]

Hope u understand...

4 0

3 years ago

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