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Leokris [45]
1 year ago
9

Minute Maid states that a bottle of juice contains 473 mL. Consumer groups are interested in determining if the bottles contain

less than the amount stated on the label. To test their claim, they sample 30 bottles. The sample mean was 472mL and the standard deviation is 0.2.
What does mu represent here?
A. The average contents of all bottles of juice in the sample, which is unknown.
B. The average contents of all bottles of juice in the population, which is unknown.
C. The average contents of all bottles of juice in the population, which is 472mL.
D. The average contents of all bottles of juice in the sample, which is 472mL.
Mathematics
1 answer:
olga2289 [7]1 year ago
6 0

Answer:

Option C) The average contents of all bottles of juice in the population, which is 473 mL

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 473 mL

Sample mean, \bar{x} = 472 mL

Sample size, n = 30

Sample standard deviation, σ = 0.2

First, we design the null and the alternate hypothesis

H_{0}: \mu = 473\text{ mL}\\H_A: \mu < 473\text{ mLinches}

Representation of \mu

  • \mu is the population parameter that represents the population mean.

Thus, for the given situation \mu represents:

Option C) The average contents of all bottles of juice in the population, which is 473 mL

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nadezda [96]

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2 years ago
Machines A and B always operate independently and at their respective constant rates. When working alone, Machine A can fill a p
pychu [463]

Answer:

The value of x is \frac{10}{3} hours.

Step-by-step explanation:

Machine A = 5 hours

Machine B = x hours

Machine A and B = 2 hours

Using the formula: \frac{T}{A}  + \frac{T}{B} = 1

where:

T is the time spend by both machine

A is the time spend by machine A

B is the time spend by machine B

\frac{2}{5}  + \frac{2}{x}  = 1

Let multiply the entire problem by the common denominator (5B)

5x(\frac{2}{5}  + \frac{2}{x} = 1)

2x + 10 = 5x

Collect the like terms

10 = 5x - 2x

10 = 3x

3x = 10

Divide both side by the coefficient of x (3)

\frac{3x}{3}  = \frac{10}{3}

x = \frac{10}{3} hours.

Therefore, Machine B will fill the same lot in \frac{10}{3} hours.

7 0
2 years ago
The graph shows the distance a ball has traveled x seconds after it was thrown what is the average speed between 2 seconds and 8
docker41 [41]

Answer:

0.5

Step-by-step explanation:

5 0
2 years ago
The point (1, 4) lies on a circle that is centered at (1, 1). Which statements are correct? Check all that apply.
Semenov [28]

the circle's radius is 3 units

the point (- 2, 1 ) lies on the circle

the equation of a circle in standard form is

(x - a)² + (y - b)² = r²

where (a, b) are the coordinates of the centre and r is the radius

the radius is the distance from the centre to the point (1, 4 ) on the circle

using (1, 4) and (1,1) in the distance formula, then

r = √(1 - 1 )² + (1 - 4)² = √(0 + 9) =√9 = 3 ⇒ r² = 9

(x - 1 )² + (y - 1 )² = 9 ← equation of circle

substitute the given points into the equation and if equation is true then they lie on the circle

(- 2, 1 ) : (- 2 - 1 )² + (1 - 1 )² = 9 + 0 = 9 ← true

Hence (- 2, 1 ) lies on the circle

(3, 3 ) : (3 - 1 )² + (3 - 1 )² = 4 + 4 = 8 ≠ 9

(3, 3 ) does not lie on the circle



6 0
2 years ago
Read 2 more answers
A study was performed on green sea turtles inhabiting a certain area.​ Time-depth recorders were deployed on 6 of the 76 capture
polet [3.4K]

Answer:

One can be 99% confident the true mean shell length lies within the above interval.

The population has a relative frequency distribution that is approximately normal.

Step-by-step explanation:

We are given that Time-depth recorders were deployed on 6 of the 76 captured turtles. These 6 turtles had a mean shell length of 51.3 cm and a standard deviation of 6.6 cm.  

The pivotal quantity for a 99% confidence interval for the true mean shell length is given by;

                    P.Q.  =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean shell length = 51.3 cm

             s = sample standard deviation = 6.6 cm

             n = sample of turtles = 6

             \mu = true mean shell length

Now, the 99% confidence interval for \mu =  \bar X \pm t_(_\frac{\alpha}{2}_)  \times \frac{s}{\sqrt{n} }

Here, \alpha = 1% so  (\frac{\alpha}{2}) = 0.5%. So, the critical value of t at 0.5% significance level and 5 (6-1) degree of freedom is 4.032.

<u>So, 99% confidence interval for</u> \mu  =  51.3 \pm 4.032 \times \frac{6.6}{\sqrt{6} }

                                                         = [51.3 - 10.864 , 51.3 + 10.864]

                                                         = [40.44 cm, 62.16 cm]

The interpretation of the above result is that we are 99% confident that the true mean shell length lie within the above interval of [40.44 cm, 62.16 cm].

The assumption about the distribution of shell lengths must be true in order for the confidence interval, part a, to be valid is that;

C. The population has a relative frequency distribution that is approximately normal.

This assumption is reasonably satisfied as the data comes from the whole 76 turtles and also we don't know about population standard deviation.

3 0
1 year ago
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