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Alina [70]
2 years ago
7

A scatterplot shows a strong, positive, linear relationship between the number of rebounds a basketball team averages and the nu

mber of wins that team records in a season. Which conclusion is most appropriate?
(A) A team that increases its number of rebounds causes its chances of winning more games to increase.
(B) If the residual plot shows no pattern, then it is safe to conclude that getting more rebounds causes more wins, on average.
(C) If the residual plot shows no pattern, then it is safe to conclude that getting more wins causes more rebounds, on average.
(D) If the r^2 value is close enough to 100%, then it is safe to conclude that getting more rebounds causes more wins, on average.
(E) Rebounds and wins are positively correlated, but we cannot conclude that getting more rebounds causes more wins, on average.
Mathematics
1 answer:
frutty [35]2 years ago
3 0

Answer: E

Conclusion E is most appropriate

Step-by-step explanation:

Rebounds and wins are positively correlated, but we cannot conclude that getting more rebounds causes more wins, on average.

Because if a scatterplot shows a strong, positive, linear relationship between two variables, then the two variables are positively correlated but there is no causation between them.

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A convex lens has a diameter of 30 mm and a depth of 5 mm at the center. How far from the vertex is the focus?
natima [27]
The formula for determining the distance of the focus from the vertex is as follows,

     f = x² / 4a

where f is focus, x is the radius (half the value of diameter), and a is the depth. Substituting the known values to the given equation,

   f = (30/2 mm)² / (4)(5 mm)

    f = 11.25 mm

<em>ANSWER: 11.25 mm</em>
8 0
2 years ago
There are 11 portable mini suites (a.k.a. cages) in a row at the Paws and Claws Holiday Pet Resort. They are neatly labeled with
BigorU [14]

Step-by-step explanation:

a) 7!

If there are no restrictions, answer is 7! as it is the permutation of all animals.

b) 4! x 3!

As cats are 6 and Dogs are 5, thus 1st and last must be cats in order to have alternate arrangements. Therefore the only choices are the order of the cats among  themselves and the order of the dogs among themselves. There are 4! permutations of the cats and 3! permutations of the dogs, so there are a total of 4! x 3! possible arrangements of the suites.

c) 3! x 5!

There are 3! possible arrangements of  the dogs among themselves. Now, if we consider the dogs as  one ”object” together, then we can think of arranging the 4 cats  together with this 1 additional object. There are 5! such arrangements possible, so there are a total of 3! · 5! possible arrangements of the suites.

d) 2 x 4! x 3!

As required that all the cats must be together and all the dogs must be together, either the cats are all  before the dogs or the dogs are all before the cats. There are two possible arrangements thus two times of both possibilities is the answer i.e.  2 x 4! x 3!

3 0
2 years ago
The perimeter of a rectangle is 13cm and its width is 11/4. find its length.
enyata [817]
Perimeter = 2 length + 2 width

P = 2(l) + 2(11/4)

13 = 2l +2* \frac{11}{4}

13 = 2l + \frac{2}{1} * \frac{11}{4}

13 = 2l + \frac{22}{4}     (subtract 22/4 from each side)

13 -\frac{22}{4} = 2l

13 -\frac{11}{2} = 2l

\frac{26}{2}  -\frac{11}{2} = 2l

\frac{15}{2}} = 2l

7.5 = 2l    (divide each side by 2)

7.5 ÷ 2 = l

3.75 = l

The answer is 3.75 or 3 \frac{3}{4}


6 0
2 years ago
In a study of the progeny of rabbits, Fibonacci (ca. 1170-ca. 1240) encountered the sequence now bearing his name. The sequence
expeople1 [14]

The question in part c is not clear, nevertheless, part a and part b would be solved.

Answer:

a. The first twelve terms are:

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144.

b. The first ten terms are:

1.000, 1.000, 1.500, 1.667, 1.600, 1.625, 1.615, 1.619, 1.618, 1.618.

Step-by-step explanation:

a. Given

an + 2 = an + an + 1

where a1 = 1 and a2 = 1.

a3 = a1 + a2

= 2

a4 = a2 + a3

= 3

a5 = a3 + a4

= 5

a6 = a5 + a4

= 8

a7 = a6 + a5

= 13

a8 = a7 + a6

= 21

a9 = a8 + a7

= 34

a10 = a9 + a8

= 55

a11 = a10 + a9

= 89

a12 = a11 + a10

= 144

The first twelve terms are:

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144.

(b)

Given

bn = an+1/an

b1 = a2/a1

= 1/1 = 1.000

b2 = a3/a2

= 2/1 = 1.000

b3 = a4/a3

= 3/2 = 1.500

b4 = a5/a4

= 5/3 = 1.667

b5 = a6/a5

= 8/5 = 1.600

b6 = a7/a6

= 13/8 = 1.625

b7 = a8/a7

= 21/13 = 1.615

b8 = a9/a8

= 34/21 = 1.619

b9 = a10/a9

= 55/34 = 1.618

b10 = a11/a10

= 89/55 = 1.618

The first ten terms are:

1.000, 1.000, 1.500, 1.667, 1.600, 1.625, 1.615, 1.619, 1.618, 1.618.

6 0
1 year ago
Which is greater 2m or 275cm
Whitepunk [10]



There is 100 centimeters in 1 meter.

Change the meter to centimeter. Multiply: 2 x 100 = 200

200 cm < 275 cm ∴  is greater by 75 cm

5 0
2 years ago
Read 2 more answers
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