here is the histogram of a data distribution. all class widths are 1. which of the following numbers is closest to the mean of this distribution a)3 b)5 c)4 d)9 e)7

2(1+4)2 x (9-6)3

4(5) x 3(3)

20 x 9

180

Answer: 600

Step-by-step explanation: i took test and got it right

Answer: 600

Step-by-step explanation:

Answer:

X=7

Step-by-step explanation:

3x+4=25 (Subtract 4 on both sides)

3x=21 (Divide by 3 on both sides)

x=7

Answer: x is a variable and the number the variable is standing for is 7

Step-by-step explanation:

3x7=21 .. 21+4=25

For the rectangle to have a perimeter of 36 units, the sum of the length and the width must equal 18 units

The given parameter is:

Perimeter = 36 units

The perimeter of a rectangle is:

Perimeter = 2 * (Length + Width)

So, we have;

36 = 2 * (Length + Width)

Divide both sides by 2

Length + Width = 18

This means that for the rectangle to have a perimeter of 36 units, the sum of the length and the width must equal 18 units

So, the possible table of values is:

Length  Width

  5           13

  8           10

  9           9

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Answer:

0.691

Step-by-step explanation:

Since a percentage means out of 100, divide the percentage by 100

69.1÷100= 0.691

0.691

100% = 1.00 helps me remember where the decimal goes.

We need to calculate the value of cos(43) to determine the value of p

The equation is given as:

cos43 = p

Rewrite properly as:

cos(43) = p

Rewrite the equation as:

p = cos(43)

The above means that we simply calculate the value of cos(43) to determine the value of p

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Answer:A) 24 waysB) 4 waysStep-by-step explanation:a) permutation occurrs when order of choices matters.N = 4P3 = 4!/(4-3)! = 4!/1!N = 24 waysb) combination occurs when order of choices doesn't matter.N = 4C3 = 4!/3!(4-3)! = 4!/3!(1!)N = 4 ways

Step-by-step explanation:

a) permutation occurrs when order of choices matters.N = 4P3 = 4!/(4-3)! = 4!/1!N = 24 waysb) combination occurs when order of choices doesn't matter.N = 4C3 = 4!/3!(4-3)! = 4!/3!(1!)N = 4 ways(hope this helps:)

Answer:

you would need 11 containers but you would have some left over.

Step-by-step explanation:

Answer:

3(3z + 5)

Step-by-step explanation:

You have to factor out 3 from 9z+15 :)

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Answer:

f(g(4)) = 13.

Step-by-step explanation:

g(4) = 4^2 = 16

so f(g(4)) = f(16)

=  16 - 3

= 13.

Answer:A

nswer is D. four over pi

Step-by-step explanation:

x1=0, y1=3

x2=pi/2, y2=5

rate of change = (y2-y1)/(x2-x1)

=(5-3)/(pi/2-0)

=4/pi

Answer:

d. [tex]a^2 =b(b+c)[/tex]

Step-by-step explanation:

Option d is the correct answer.

Answer: you will get 45

Step-by-step explanation:

If you do 9x5=45 and 11x5=55

So the answer would be 45

Answer:

the answer is bottom left trust

Answer:

bottom left is the answer

Step-by-step explanation:

5/6 or 0.83 boxes of raisins

Equation of circle in general

For centre at (h,k)

So our equation is

Answer:

1000

Step-by-step explanation:

Answer: y = (x/2)°

Step-by-step explanation:

The two angles are because they are opposite angles.

Answer:

The distance between the points (2, 7) and (2, 15) is 8.

Step-by-step explanation:

What is the Distance between Two Points?

For any two points there is exactly one line segment connecting them. The distance between two points is the length of the line segment connecting them. Note that the distance between two points is always positive. Segments that have equal length are called congruent segments.

How to Solve:

Putting Values:

[tex]=\sqrt{(2-2)^2+(15-7)^2}[/tex]

[tex]=\sqrt{(0)^2+(8)^2}[/tex]

[tex]=\sqrt{0+64}[/tex]

[tex]=\sqrt{64}[/tex]

[tex]=8[/tex]

Answer:

Yes, by ASA congruence.

Step-by-step explanation:

Solving :

Hence, as 2 angles and a side of the triangles are equal, we can say they are congruent by ASA congruence.

[tex]\huge\underline{{\sf Answer}}[/tex]

[tex] \textbf{Let's check if the two triangles are } [/tex][tex] \textbf{congruent} [/tex] ~

[tex] \large\textsf{Given information } [/tex]:

[tex] \textsf{Two angles of a triangle is equal to corresponding} [/tex] [tex] \textsf{two angles of another triangle, and they} [/tex][tex] \textsf{have one side in common (should be equal} [/tex][tex] \textsf{for both) } [/tex]

[tex] \textsf{So, after analyzing this information, we can } [/tex][tex] \textsf{conclude that the two triangles are congruent by} [/tex][tex] \textsf{AAS congruency criteria.} [/tex]

[tex] \texttt{[ since two angles are equal and one} [/tex][tex] \texttt{non included side is common/equal ]} [/tex]

Answer:

first solve is 10 nickels and 1 dime

second solve is 4 quarters and 7 dimes

Step-by-step explanation:

see attached picture of both math details

Answer:

10 duh

Step-by-step explanation:

Step-by-step explanation:

answer is in the picture above

Answer:

21 mm

Step-by-step explanation:

Given,

Volume of cone ( V ) = 2307.9 mm³

Height ( h ) = 5 mm

To find : Radius ( r )

Formula : -

Volume of cone = πh/3 r²

Note : -

The value of π is 3.14

Volume of cone = πh/3 r²

2307.9 mm³ = 3.14 x 5/3 x r²

2307.9 = 5.23 x r²

Divide 5.23 on both sides,

r² = 2307.9 / 5.23

r² = 441

r² = 21 x 21

r² = 21²

r = 21 mm

Answer:

The answer would be fx=4(2x)

Step-by-step explanation:

have a good day!!

The total number of different committees that can be made is:

C = 4,536

If we have a set of N elements, the total number of different subsets of K elements that can be made out of these N is:

[tex]C(N, K) = \frac{N!}{(N - K)!*K!}[/tex]

So, the 2 teachers (out of 9) can be selected in:

[tex]C(9, 2) = \frac{9!}{7!*2!} = \frac{9*8}{2} = 36[/tex]

36 different ways.

The 4 students (out of 9) can be selected in:

[tex]C(9, 4) = \frac{9!}{5!*4!} = \frac{9*8*7*6}{4*3*2} = 126[/tex]

126 different ways.

The committee is formed by these two groups, so the total number of different committees that can be made is:

C = 36*126 = 4,536

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[tex]\qquad \qquad \textit{direct proportional variation} \\\\ \textit{\underline{y} varies directly with \underline{x}}\qquad \qquad \stackrel{\textit{constant of variation}}{y=\stackrel{\downarrow }{k}x~\hfill } \\\\ \textit{\underline{x} varies directly with }\underline{z^5}\qquad \qquad \stackrel{\textit{constant of variation}}{x=\stackrel{\downarrow }{k}z^5~\hfill } \\\\[-0.35em] ~\dotfill[/tex]

[tex]\stackrel{\textit{"y" varies directly with "x"}}{y=kx}\qquad \textit{we also know that} \begin{cases} x=\frac{2}{3}\\ y = 6 \end{cases}\implies 6=k\cfrac{2}{3} \\\\\\ 18=2k\implies \cfrac{18}{2}=k\implies 9=k~\hfill \boxed{y=9x} \\\\\\ when~x=10\frac{1}{3}\textit{ what is "y"?}\qquad y=9\left( 10\frac{1}{3} \right)\implies y=9\cdot \cfrac{31}{3}\implies y=93[/tex]