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jeka57 [31]
2 years ago
8

The Atlantic bluefin tuna is the largest and most endangered of the tuna species. They are found throughout the North Atlantic O

cean and the mean weight is 550 pounds.37 The Center for Biological Diversity is concerned that this species has been overfished and that the mean weight has decreased. Suppose a random sample of 12 Atlantic bluefin tuna was obtained from commercial fishing boats and weighed. The summary statistics were x 5 535.7 and s 5 37.8. Conduct a hypothesis test to determine whether there is any evidence that the mean weight of Atlantic bluefin tuna is less than 550 pounds. Assume the distribution of weight is normal and use a 5 0.05. Write a Solution Trail for this problem.

Mathematics
1 answer:
Maksim231197 [3]2 years ago
3 0

Answer:

Please see attachment

Step-by-step explanation:

Please see attachment

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Consider the graph of the line y = x – 4 and the point (−4, 2). The slope of a line parallel to the given line is . A point on t
Harman [31]
The given line is y = x - 4, and the given point is (-4, 2).

The slope of the given line is 1, and its y-intercept is (0, -4).

Part a.
The slope of a line parallel to the given line is 1.

Part b.
Let the equation of a line parallel to the given line be
y = x + c
Let (x,y) be a point on a parallel line which passes through (-4,2).
Then, in point-slope form,
(y - 2)/(x + 4) = 1
That is,
y - 2 = x + 4
y = x + 6

If x = 0, then
y - 2 = 0 + 4
y - 2 = 4
y = 6
Therefore the point (0, 6)  is one of many points that lies on this line.

Part c.
The slope of a line perpendicular to the given line is -1.
The product of the slopes of perpendicular lines is -1.

Let (x,y) be a point on the perpendicular line which passes through (-4,2).
In point-slope form, its equation is
(y - 2)/(x + 4) = -1
That is,
y - 2 = -x - 4
y = -x - 2
Wneh x = 0, then y = -2.
A point on the perpendicular line that passes through (-4,2) is (0, -2).

Answers:
1. The slope is 1.
2. The point (0, 6) falls on the parallel line passing through (-4, 2).
3. The slope of a perpendicular line is  -1.
4. The point (0, -2) falls on the perpendicular line passing through (-4, 2).

8 0
2 years ago
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A large school district is considering moving up the start date for the school year by two weeks. In order to determine if famil
Anvisha [2.4K]

Answer:

(a) Confirmation bias

(b) Likely less than actual proportion of families who support the proposal

(c) Random sample

(d) Nonresponse bias

Step-by-step explanation:

In confirmation bias, the researcher already has and intends to prove an assumption about the effect the of a treatment before carrying out the survey, such that the survey or research is tries to dictate to the participants in the survey about the desired outcome

In the question, the belief of the researcher is expressed to the participants in a form of brief at the start of the survey which is attempting to influence the responses to the survey, such that the actual result should be less than what was observed.

5 0
1 year ago
I tell you these facts about a mystery number, $c$: $\bullet$ $1.5 < c < 2$ $\bullet$ $c$ can be written as a fraction wit
makkiz [27]

Answer:

Possible answer: \displaystyle c = \frac{16}{10} = \frac{8}{5} = 1.6.

Step-by-step explanation:

Rewrite the bounds of c as fractions:

The simplest fraction for 1.5 is \displaystyle \frac{3}{2}. Write the upper bound 2 as a fraction with the same denominator:

\displaystyle 2 = 2 \times 1 = 2 \times \frac{2}{2} = \frac{4}{2}.

Hence the range for c would be:

\displaystyle \frac{3}{2} < c < \frac{4}{2}.

If the denominator of c is also 2, then the range for its numerator (call it p) would be 3 < p < 4. Apparently, no whole number could fit into this interval. The reason is that the interval is open, and the difference between the bounds is less than 2.

To solve this problem, consider scaling up the denominator. To make sure that the numerator of the bounds are still whole numbers, multiply both the numerator and the denominator by a whole number (for example, 2.)

\displaystyle \frac{3}{2} = \frac{2 \times 3}{2 \times 2} = \frac{6}{4}.

\displaystyle \frac{4}{2} = \frac{2\times 4}{2 \times 2} = \frac{8}{4}.

At this point, the difference between the numerators is now 2. That allows a number (7 in this case) to fit between the bounds. However, \displaystyle \frac{1}{c} = \frac{4}{7} can't be written as finite decimals.

Try multiplying the numerator and the denominator by a different number.

\displaystyle \frac{3}{2} = \frac{3 \times 3}{3 \times 2} = \frac{9}{6}.

\displaystyle \frac{4}{2} = \frac{3\times 4}{3 \times 2} = \frac{12}{6}.

\displaystyle \frac{3}{2} = \frac{4 \times 3}{4 \times 2} = \frac{12}{8}.

\displaystyle \frac{4}{2} = \frac{4\times 4}{4 \times 2} = \frac{16}{8}.

\displaystyle \frac{3}{2} = \frac{5 \times 3}{5 \times 2} = \frac{15}{10}.

\displaystyle \frac{4}{2} = \frac{5\times 4}{5 \times 2} = \frac{20}{10}.

It is important to note that some expressions for c can be simplified. For example, \displaystyle \frac{16}{10} = \frac{2 \times 8}{2 \times 5} = \frac{8}{5} because of the common factor 2.

Apparently \displaystyle c = \frac{16}{10} = \frac{8}{5} works. c = 1.6 while \displaystyle \frac{1}{c} = \frac{5}{8} = 0.625.

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It will last around 10 days
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a straight highway is 100 mile long and each mile is marked by a milepost numbered from 0 to 100. a rest area is going to be bui
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