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iren [92.7K]
2 years ago
7

I need help solving this pls

Mathematics
1 answer:
Oksana_A [137]2 years ago
8 0

Answer:

Adding them together. x = 3/2, y = 8

Step-by-step explanation:

Adding because we want to get rid of a variable. If we add, 2x + 6x = 8x, 1/2 * y + (-1/2 * y) = 0, and 7 + 5 = 12. Adding gets rid of the y variable.

By adding we get that 8x = 12 which gives us x = 12/8 = 3/2

Substituting this back into either equation will give us y = 8.

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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
2 years ago
A certain company's main source of income is selling socks. The company's annual profit (in millions of dollars) as a function o
ivann1987 [24]
A quadratic function models the profit of the company and the maximum/minimum value of the quadratic function occurs on its vertex. 

The given function is:

P(x)=-3(x-5)^{2}+12

The equation is already in standard form, the vertex of the parabola as seen from the equation is (5,12)

This mean if the company sets the price of socks to $5 they will earn a maximum profit which is $ 12 million
3 0
3 years ago
Read 2 more answers
Round 26.55 to the nearest whole number
Natasha_Volkova [10]

The next larger whole number is  27 .
The next smaller whole number is  26 .

26.55 is closer to  27  than it is to  26 .

So the nearest whole number is  27 .

3 0
3 years ago
Read 2 more answers
Question 1 of 5
Murljashka [212]

Answer:

the equation would be b. y=2x-3.

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2 years ago
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The points (3, 9) and (–3, –9) are plotted on the coordinate plane using the equation y = a • x.
marysya [2.9K]
The answer would be a=3
7 0
2 years ago
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