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lakkis [162]
3 years ago
14

Two glasses of milk and 4 snack bars have a total of 66 carbohydrates​ (carbs), and 4 glasses of milk and 3 snack bars have a to

tal of 82 carbs. Determine how many carbs are in one glass of milk and in one snack bar
Mathematics
1 answer:
musickatia [10]3 years ago
5 0
2m + 4s = 66.
4m + 3s = 84.
(Double the 2 glass of milk and 4 snack bars)
4m + 8s = 132.
4m + 3s = 84.
(Take away the second equation from the first equation)
5s= 48.
s = 48/5.
s = 9.6.
4m + 3(9.6) = 84.
4m + 28.8 = 84.
4m = 84 - 28.8.
4m = 55.2.
4m/4 = 55.2/4.
m = 13.8.
In a glass of milk, there are 13.8 carbs, and in a snack bar, there are 9.6 carbs.
2(13.8) + 4(9.6) = 66.
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Step-by-step explanation:

We are given that a laboratory scale is known to have a standard deviation (sigma) or 0.001 g in repeated weighing. Scale readings in repeated weighing are Normally distributed with mean equal to the true weight of the specimen.

Three weighing of a specimen on this scale give 3.412, 3.416, and 3.414 g.

Firstly, the pivotal quantity for 99% confidence interval for the true mean specimen is given by;

        P.Q. = \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \bar X = sample mean weighing of specimen = \frac{3.412+3.416+3.414}{3} = 3.414 g

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            n = sample of specimen = 3

            \mu = population mean

<em>Here for constructing 99% confidence interval we have used z statistics because we know about population standard deviation (sigma).</em>

So, 99% confidence interval for the population​ mean, \mu is ;

P(-2.5758 < N(0,1) < 2.5758) = 0.99  {As the critical value of z at 0.5% level

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P(-2.5758 < \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } < 2.5758) = 0.99

P( -2.5758 \times {\frac{\sigma}{\sqrt{n} } } < {\bar X - \mu} < 2.5758 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.99

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