The table displays the scores of students on a recent exam. Find the mean of

1 answer:

5 0

The answer is 44.5. You would add all of the numbers shown above. This would get you a sum of 623. You would then count how many numbers there are in the question which is 14. Then divide 623 and 14 to get an average if 44.5

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The sum of three consecutive even numbers 86. Select all the equations below that represent this situation.

Elena-2011 [213]

Answer:
A = X+X+2+X+4=3X+6 =(X+1)+(X+2)+(X+3)
B = 3X+6 =(X+1)+(X+2)+(X+3)
C = not (3X+3)
D = not (3X+3)

3 0

3 years ago

Fraction of the shape shaded on these two shapes??

xxTIMURxx [149]

The one of the left isn't too tricky, in the bottom left of it you have 1/2 of 1/4 of the shape, and on the bottom right you have 1/4 of 1/4. Imagine the whole shape was cut up like that bottom right bit into 16 triangles, then you would have three of them shaded (see my diagram).

The one on the right seems like guesswork to me, so I'm sorry if I'm wrong. It look like you just have to use your eyes to work out how much of the bottom half of the shape is shaded: looks like 1/16 to me (i.e. you can draw four of those along and four down, again, see my diagram.) So plus the top half, which is 8/16, you get 9/16.

Answers: left picture: 3/16.
right picture: 9/16.

3 0

3 years ago

Plz help me with this

AnnZ [28]

Answer: $\bold{A)\quad y=5sin\bigg(\dfrac{6}{5}x-\pi\bigg)-4}$

<u>Step-by-step explanation:</u>

The general form of a sin/cos function is: y = A sin/cos (Bx-C) + D where the period (P) = 2π ÷ B

In the given function, $B=\dfrac{6}{5}$ → $P=2\pi \cdot \dfrac{5}{6}=\dfrac{10\pi}{3}$

Half of that period is: $\dfrac{1}{2}\cdot \dfrac{10\pi}{3}=\large\boxed{\dfrac{5\pi}{3}}$

Calculate the period for each of the options to find a match:

$A)\quad B=\dfrac{6}{5}:\quad 2\pi \div \dfrac{6}{5}=2\pi \cdot \dfrac{5}{6}=\dfrac{5\pi}{3}\quad \leftarrow\text{THIS WORKS!}\\\\\\B)\quad B=\dfrac{6}{10}:\quad 2\pi \div \dfrac{6}{10}=2\pi \cdot \dfrac{10}{6}=\dfrac{10\pi}{3}\\\\\\C)\quad B=\dfrac{5}{6}:\quad 2\pi \div \dfrac{5}{6}=2\pi \cdot \dfrac{6}{5}=\dfrac{12\pi}{5}\\\\\\D)\quad B=\dfrac{3}{10}:\quad 2\pi \div \dfrac{3}{10}=2\pi \cdot \dfrac{10}{3}=\dfrac{20\pi}{3}$