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3 years ago

Please help on this equation: -2(-w+3)=15, please show work. thank you

1 answer:

3 0

-w is the variable, so our goal is to get it on the left side of the equal sign and keep the 15 on the right side.
distribute the -2 to each term inside the parentheses. the equation should now look like this:
2w - 6 = 15
now, we must transpose the -6 to the other side of the equation. remember, we're moving it across the equal sign, so it becomes a positive 6.
2w = 21
now divide each side of the equation by 2 so the variable is alone.
w = 10.5
hope i helped!

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Factor the expression below x^2+12+36 A) (x-6)(x-6) B) (x+6)(x+6) C) (x-9)(x-4) D) (x+9)(x+4)

evablogger [386]

Answer:

Answer B

Step-by-step explanation:

when you simplify it u get x^2+12+36

4 0

3 years ago

How do you subtract unlike fractions?

nasty-shy [4]

You find the LCM of the denominators in the fractions, then you multiply both fractions as needed till both have like denominators. Then you subtract.

For example:

5/9 - 2/4

LCM of 9 and 4 is 36

[(5 x 4)/(9 x 4)] - [(2 x 9) - (4 x 9)]
= 20/36 - 18/36
20/36 - 18/36 = 2/36
2/36 = 1/18

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5 0

3 years ago

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

Use the inequalities to describe the shaded area on the grid

raketka [301]

The shaded area would be 2 because there is a whole square shaded in the the other two are 1/2 so if you add 1/2 together with another 1/2 plus the whole shaded square you would get two.

At least that’s what I understood from the question

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3 years ago

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kobusy [5.1K]

Answer:

i think the answer is D my guy

Step-by-step explanation:

do you need the steps??

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3 years ago