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3 years ago

What is the value of the interquartile range of the data below? 6 12 14 24

Mathematics

1 answer:

Strike441 [17]3 years ago

7 0

Answer:

Step-by-step explanation:

The interquartile range is the value of quartile 3 minus quartile 1.

The "box" part of this diagram has 3 lines, the left most, the middle, and the rightmost.

The leftmost line is Quartile 1. The right most is Quartile 3.

Hence

interquartile range = quartile 3 - quartile 1

Looking at the box plot, we can see that each small line in the number line is 2 units.

The leftmost line (quartile 1 ) is at 1 unit left of 30, so that is 30 -2 = 28

The rightmost line (quartile 3) is at 1 unit right of 40, so that is 40 + 2 = 42

Hence,

Interquartile range = 42 - 28 = 14

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Answer:

Σ(-1)^kx^k for k = 0 to n

Step-by-step explanation:

The nth Maclaurin polynomials for f to be

Pn(x) = f(0) + f'(0)x + f''(0)x²/2! + f"'(0)x³/3! +. ......

The given function is.

f(x) = 1/(1+x)

Differentiate four times with respect to x

f(x) = 1/(1+x)

f'(x) = -1/(1+x)²

f''(x) = 2/(1+x)³

f'''(x) = -6/(1+x)⁴

f''''(x) = 24/(1+x)^5

To calculate with a coefficient of 1

f(0) = 1

f'(0) = -1

f''(0) = 2

f'''(0) = -6

f''''(0) = 24

Findinf Pn(x) for n = 0 to 4.

Po(x) = 1

P1(x) = 1 - x

P2(x) = 1 - x + x²

P3(x) = 1 - x+ x² - x³

P4(x) = 1 - x+ x² - x³+ x⁴

Hence, the nth Maclaurin polynomials is

1 - x+ x² - x³+ x⁴ +.......+(-1)^nx^n

= Σ(-1)^kx^k for k = 0 to n

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3 years ago

A w=-8<br> B w=-5 <br> C w=5<br> D w=8

algol [13]

Answer:

Step-by-step explanation:

Plug in every scenario, in see if it equal to each other.

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3 years ago

Find the equation of the line that passes through the points (2, 2) and (-4, -1).

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The answer is c. if you plug in the numbers for x you get the correct numbers for y

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Step-by-step explanation:

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A group of 30 students from your school is part of the audience for a TV game show. The total number of people in the audience i

s344n2d4d5 [400]

Answer:

5.3%

Step-by-step explanation:

The final probability is calculated by means of the quotient of the specific combinations and the total of total combinations

Let's start with the specific ones,

First the number of combinations of 4 of the 30 students getting a spot, i.e .:

A combinations are equal to:

nCx = n! / x! * (n-x)!

Replacing:

30C4 = 30! / (4! * 26!) = 27405

Segundo the number of combinations of the other audience members filling the other 4 (8-4) spots n = 110, 140 - 30

110C4 = 110! / (4! * 106!) = 5773185

Now the total combinations of possible 8 contestants from the audience

140C8 = 140! / (8! * 132!) = 2.98 * 10 ^ 12

Finally, the probability is equal to:

P = (30C4 * 110C4) / 140C8

replacing:

P = 27405 * 5773185 / 2.98 * 10 ^ 12

P = 0.053

Therefore the probability is 5.3%

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3 years ago