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sammy [17]
3 years ago
13

Identify any values of data that might affect the statistical measures of spread and center.

Mathematics
1 answer:
monitta3 years ago
5 0
Answer: There is a high data value that causes the data set to be asymmetrical for the males.

The females 'might' have an outlier, but it's difficult to say exactly.

An outlier is defined as any data value greater than 1.5 times the Interquartile range or less.
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What is the value of the expression when n = 3?<br> GO<br> - 2 n(5 + n-8-3n)
lukranit [14]

Answer:

54

Step-by-step explanation:

since n = 3

= -2n ( 5n+n-8-3n)

= -2 (3) ( 5+(3)-8-3(3) )

= -6 (5+3-8-9)

= -6 (5-5-9)

= -6 (-9)

= 54.

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3 years ago
What is 14% of $59.99 and how do I figure this problem?
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Graph from liner standard form 3x+2y=12?​
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Find the exact value of the remaining trigonometric functions given: Tan x = Undefined Sin x &gt; 0
finlep [7]

Answer:

  • sin(x) = 1
  • cos(x) = 0
  • cot(x) = 0
  • csc(x) = 1
  • sec(x) = undefined

Step-by-step explanation:

The tangent function can be considered to be the ratio of the sine and cosine functions:

  tan(x) = sin(x)/cos(x)

It will be undefined where cos(x) = 0. The values of x where that occurs are odd multiples of π. The smallest such multiple is x=π/2. The value of the sine function there is positive: sin(π/2) = 1.

The corresponding trig function values are ...

  tan(x) = undefined (where sin(x) >0)

  sin(x) = 1

  cos(x) = 0

__

And the reciprocal function values at x=π/2 are ...

  cot(x) = 0 . . . . . . 1/tan(x)

  csc(x) = 1 . . . . . . .1/sin(x)

  sec(x) = undefined . . . . . 1/cos(x)

3 0
2 years ago
About how far apart do aesha and Josh live​
Vesnalui [34]

Answer:

D. about 8.5 mi

Step-by-step explanation:

To go from Aesha to Josh, you go 6 units right and 6 units up.

Each unit is a mile, so you go 6 miles right and 6 miles up.

Think of each 6 mile distance as a leg of a right triangle, and the direct distance from one place to the other as the hypotenuse of the right triangle. Use the Pythagorean theorem to find the length of the hypotenuse.

a^2 + b^2 = c^2

The 6-mile legs are a and b. c is the hypotenuse.

(6 mi)^2 + (6 mi)^2 = c^2

c^2 = 36 mi^2 + 36 mi^2

c^2 = 72 mi^2

c = sqrt(72) mi

c = sqrt(36 * 2) mi

c = 6sqrt(2) mi

c = 6(1.4142) mi

c = 8.5 mi

8 0
3 years ago
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