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3 years ago

10

Invent a five-value data set for which the mean is 6 and the mode is not 6

1 answer:

GaryK [48]3 years ago

3 0

Mean is the average
mode is most common number
we can do
hmm
average of 3 and 9 is 6

3,3,6,9,9 will work
we can make it vary slightly
2,4,6,9,9
mean is 6, mode is 9

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Two sides of an isosceles triangle measure 3 inches and 7 inches. Which could be the length of the third side?

MrRa [10]

7? An isosceles triangle has two indentical sides(you can remember that by saying you have TWO eyes hint EYEsosceles) but I’m guessing the bottom is 3 and the two sides are 7?

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3 years ago

Each unit on the grid represents 5

ollegr [7]

Answer:

There's no grid but I'd say A if you can give me the grid I'll edit my answer

Step-by-step explanation:

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3 years ago

In△XYZ , XY¯¯¯¯¯¯=7in , YZ¯¯¯¯¯=5in , and XZ¯¯¯¯¯=4in . This triangle is reflected across the x-axis to result in △X'Y'Z' . Whic

ki77a [65]

Answer:

B

Step-by-step explanation:

Reflection is a transformation that preserves lengths.

Triangle XYZ, with side lengths XY = 7 in, YZ = 5 in, XZ = 4 in, is reflected across the x-axis to result in △X'Y'Z'.

This means that corresponding sides have the same lengths:

XY = X'Y' = 7 in;
XZ = X'Z' = 4 in;
YZ = Y'Z' = 5 in.

Find the perimeters of both triangles:

$P_{XYZ}=XY+XZ+YZ=7+4+5=16\ in\\ \\P_{X'Y'Z'}=X'Y'+X'Z'+Y'Z'=7+4+5=16\ in$

Hence, only option B is true.

8 0

3 years ago

(x2 + 10x + 21) - (x + 7)

rusak2 [61]

Answer:

11x+14

Step-by-step explanation:

(2x+10x+21)-x-7 = 12x+21-x-7 = 11x+14

3 0

3 years ago

A cylinder has a height of 8.1 inches and circular base with a diameter of 2.7 inches. The cylinder contains 3 green spheres, ea

klasskru [66]

Answer:

A point is chosen at random inside the cylinder, then probability the point belongs outside of the spheres is 0.25

Step-by-step explanation:

Let's find volume of the cylinder and volume of the three spheres.

Volume of cylinder = $\pi r^{2} h$

Volume of sphere = $\frac{4}{3} \pi r^{3}$

Now, r =2.7

h=8.1

Find volume of each shape.

Volume of cylinder = $\pi (2.7)^{2} (8.1)$

=59.049 $\pi$

Now, volume of 3 spheres = $3*\frac{4}{3} *\pi *(2.7)^{3}$

=78.732 $\pi$

Now, p(E)= $\frac{Outcomes favorable to event E}{Total number of outcomes}$

So, Probability of point is chosen at random inside the cylinder which belongs outside of the spheres is

$\frac{78.732\pi -59.049\pi }{78.732\pi }$

Let's simplify them

0.25

A point is chosen at random inside the cylinder, then probability the point belongs outside of the spheres is 0.25

6 0

3 years ago