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Elza [17]
3 years ago
11

Is x=2 a solution to the equation below? 4(x-1)-3(x-2)=-8

Mathematics
1 answer:
Sonbull [250]3 years ago
7 0

Answer:

No

Step-by-step explanation:

To find out if 2 is a solution, replace all the x's with 2.

4(2 - 1) - 3(2 - 2) =8

4(1) - 3(0) = 8

4 - 0 = 8

4 = 8

Nope!  False!  2 is not a solution.

You might be interested in
Measurements of the sodium content in samples of two brands of chocolate bar yield the following results (in grams):
Tpy6a [65]

Answer:

98% confidence interval for the difference μX−μY = [ 0.697 , 7.303 ] .

Step-by-step explanation:

We are give the data of Measurements of the sodium content in samples of two brands of chocolate bar (in grams) below;

Brand A : 34.36, 31.26, 37.36, 28.52, 33.14, 32.74, 34.34, 34.33, 29.95

Brand B : 41.08, 38.22, 39.59, 38.82, 36.24, 37.73, 35.03, 39.22, 34.13, 34.33, 34.98, 29.64, 40.60

Also, \mu_X represent the population mean for Brand B and let \mu_Y represent the population mean for Brand A.

Since, we know nothing about the population standard deviation so the pivotal quantity used here for finding confidence interval is;

        P.Q. = \frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}  } } ~ t_n__1+n_2-2

where, Xbar = Sample mean for Brand B data = 36.9

            Ybar = Sample mean for Brand A data = 32.9

              n_1  = Sample size for Brand B data = 13

              n_2 = Sample size for Brand A data = 9

              s_p = \sqrt{\frac{(n_1-1)s_X^{2}+(n_2-1)s_Y^{2}  }{n_1+n_2-2} } = \sqrt{\frac{(13-1)*10.4+(9-1)*7.1 }{13+9-2} } = 3.013

Here, s^{2}_X and s^{2} _Y are sample variance of Brand B and Brand A data respectively.

So, 98% confidence interval for the difference μX−μY is given by;

P(-2.528 < t_2_0 < 2.528) = 0.98

P(-2.528 < \frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}  } } < 2.528) = 0.98

P(-2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} < (Xbar -Ybar) -(\mu_X-\mu_Y) < 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} ) = 0.98

P( (Xbar - Ybar) - 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} < (\mu_X-\mu_Y) < (Xbar - Ybar) + 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} ) = 0.98

98% Confidence interval for μX−μY =

[ (Xbar - Ybar) - 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} , (Xbar - Ybar) + 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} ]

[ (36.9 - 32.9)-2.528*3.013\sqrt{\frac{1}{13} +\frac{1}{9} , (36.9 - 32.9)+2.528*3.013\sqrt{\frac{1}{13} +\frac{1}{9} ]

[ 0.697 , 7.303 ]

Therefore, 98% confidence interval for the difference μX−μY is [ 0.697 , 7.303 ] .

                     

4 0
3 years ago
OK you look like your going through a tuff time, the cold harden truth is that you have no choice but to live with it. Listen cl
ra1l [238]

Answer:

ok

Step-by-step explanation:

7 0
3 years ago
-2
Elodia [21]

Answer:

It would be the 2nd one

Step-by-step explanation:

4 0
3 years ago
What is the area of the polygon in square units?
Pepsi [2]

The area of the Area  of the polygon is known to be 9 units². Find more about polygon below

<h3>What is the Area of a Polygon?</h3>

The area of a polygon is one that is defined as the measurement of the given area that is known to be covered by it.

Note that to calculate the area, it will be

Area = 1/2 (5 + 1) x 3 = 9 units²

Thus the answer of 9 units is correct.

Learn more about polygon from

brainly.com/question/1592456

#SPJ1

4 0
2 years ago
Please help with algebra 2 logarithms. brainliest if you explain and worth 25 points
Ronch [10]

Answer: 3log5(2) - 1 or ~0,29203

Step-by-step explanation:

2log5(4) - log5(10)

log5(4^2) - log5(10)

log5(4^2/10)

log5(16/10)

log5(8/5)

Log5(8) - log5(5)

log5(2^3) - 1

3log5(2) - 1

5 0
2 years ago
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