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Harrizon [31]
3 years ago
13

4y = 42 - 3y Can you help me find y?

Mathematics
2 answers:
Lesechka [4]3 years ago
7 0
4y = 42 - 3y

First, add '3y' to both of the sides.
4y + 3y = 42
Second, add '4y + 3y' to get '7y'.
7y = 42
Third, divide both sides by '7'.
y =  \frac{42}{7}
Fourth, how many times does 7 go into 42? 42 ÷ 7 = '6'.
y = 6

Answer: y = 6

AleksAgata [21]3 years ago
6 0
Bring all the variables with y to one side:
4y = 42 - 3y
4y + 3y = 42 - 3y + 3y
7y = 42

Divide both sides by 7 to get the y alone:
7y ÷ 7 = 42 ÷ 7
y = 6
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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 ] .

                     

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An insurance agent has appointments with four prospective clients tomorrow. From past experience the agent knows that the probab
larisa86 [58]

Answer:

0.082

Step-by-step explanation:

Total number of clients = 4

From past experience, the agent knows that the probability of making a sale on any appointment is one out of five = 1/5.

This shows that there are only two possible outcomes, success or failure. It is either she makes a sale on an appointment or she fails to make a sale on the appointment. Therefore, we can use binomial distribution.

For binomial distribution,

P(x=r) = nCr × q^(n-r) × p^r

p = probability of making a sale on any appointment = 1/5 = 0.2

q = probability of not making a sale on any appointment = 1 -1/5 = 0.8

n = 5

x = 3

P(x=3) = 5C3 × 0.8^2 × 0.2^3

= 10×0.6724× 0.012167 =0.082

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katen-ka-za [31]

Answer: Infinitely Many Solutions

Step-by-step explanation:

Substitute y into the second equation:

2(-2x-6)=-4x-12

Then you get -4x-12=-4x=-12. Since this is 0=0 it would make it infinitely many solutions.

6 0
3 years ago
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