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Afina-wow [57]
3 years ago
15

Dwayne buys 1 loaf of bread and 6 eggs for $4.45. Sheila buys 1 loaf of bread and 1 dozen eggs for $5.65. Let x be the cost, in

dollars, of a loaf of bread. Let y be the cost, in dollars, of an egg.
Mathematics
1 answer:
kobusy [5.1K]3 years ago
6 0

Answer:

Kindly check explanation

Step-by-step explanation:

Given the following :

Let x = cost of a loaf of bread ; y = cost of an egg

Dwayne :

1 loaf of bread and 6 eggs :

X + 6y = 4.45 - - - (1)

Sheila :

X + 12y = 5.65 - - - (2)

From (1);

X = 4.45 - 6y

In (2):

(4.45 - 6y) + 12y = 5.65

4.45 + 6y = 5.65

6y = 5.65 - 4.45

6y = 1.2

y = 0.2

From X = 4.45 - 6y:

X = 4.45 - 6(0.2)

X = 4.45 - 1.2

X = 3.25

Cost of a loaf of bread = $3.25

Coat of an egg = $0.2

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Answer:

Hello!!!

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Use the rewritten formula to find the width of a rectangle with an area of 42 square inches and a length of 16.8 inches.
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3 years ago
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A laboratory scale is known to have a standard deviation (sigma) or 0.001 g in repeated weighings. Scale readings in repeated we
weqwewe [10]

Answer:

99% confidence interval for the given specimen is [3.4125 , 3.4155].

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

            \sigma = population standard deviation = 0.001 g

            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

                                                            of significance are -2.5758 & 2.5758}

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

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

<u>99% confidence interval for</u> \mu = [ \bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } } , \bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } } ]

                                             = [ 3.414-2.5758 \times {\frac{0.001}{\sqrt{3} } } , 3.414+2.5758 \times {\frac{0.001}{\sqrt{3} } } ]

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Therefore, 99% confidence interval for this specimen is [3.4125 , 3.4155].

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Answer:

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