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Aleks04 [339]
2 years ago
5

Please help me .... Thank you ​

Mathematics
1 answer:
Verizon [17]2 years ago
3 0

\\ \sf\longmapsto \dfrac{4}{6x+2}

\\ \sf\longmapsto \dfrac{5x+3y}{16-8y}

\\ \sf\longmapsto \dfrac{2}{4x-3}

\\ \sf\longmapsto \dfrac{2ab+c^2}{3n-2m}

\\ \sf\longmapsto \dfrac{a^2-16}{7x}

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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
Rewrite the equation <br> x + y + 3 = -2x + 4y + 18 <br> into general form
dolphi86 [110]

Answer:

x-y=5 just solve it to get the answer

5 0
2 years ago
(URGENT)Find the equation of the normal line to the curve below at the point (1, -2) in point slope form.
Rus_ich [418]

Answer:

Xmin:

-10

Xmax:

10

Ymin:

-10

Ymax

10

Step-by-step explanation:

4 0
3 years ago
What is the net force acting on a 3000kg sled that accelerated a 5 m/s2
katovenus [111]

Answer:

15000 Newtons

Step-by-step explanation:

Force (in newtons)= weight * acceleration = 3000kg * 5 m/s^2 = 15000 N

5 0
3 years ago
Miranda will have 500 guests at her Sweet Sixteen party. She conducted a random survey of 50 guests to determine how many of eac
antiseptic1488 [7]

Answer:

240 sodas, 160 water, 100 juice

Step-by-step explanation:

multiply everything by 10

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2 years ago
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