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3 years ago

7

A gum ball machine has 6 red gum balls, 3 green gum balls, and 5 blue gum balls. What is the probability that the next gum ball

that comes out will be red?

1 answer:

lord [1]3 years ago

7 0

Answer:

Step-by-step explanation:

P(red gum ball)= 6/14=3/7

6/14 can be simplified to 3/7 by dividing by 2. 6/2=3. 14/2=7.

P(red gum ball)=3/7

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Answer:

20

Step-by-step explanation:

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3 years ago

Is the following graph<br> an example of a linear<br> function?<br> Yes<br> No

oksian1 [2.3K]

Answer:

no

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3 years ago

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3 years ago

What is the least common multiple of 6 and 17?<br>

amm1812

Answer:

102

Step-by-step explanation:

The least common multiple of 6 and 17

For us to determine the L.C.M of 6 and 17,

First find the prime factors of 6 and 17

Prime Factor of 6

6 = 2 × 3

Prime factor of 17

17 = 1 × 17

Secondly;

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L.C.M = 1 × 2 × 3 × 17

L.C.M = 102

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6 0

3 years ago

The height h (in feet) of an object dropped from a ledge after x seconds can be modeled by h(x)=−16x2+36 . The object is dropped

kakasveta [241]

Check the picture below.

$\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(t) = -16t^2+v_ot+h_o \quad \begin{cases} v_o=\textit{initial velocity}&\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&\\ \qquad \textit{at "t" seconds} \end{cases}$

so the object hits the ground when h(x) = 0, hmmm how long did it take to hit the ground the first time anyway?

$\bf h(x)=-16x^2+36\implies \stackrel{h(x)}{0}=-16x^2+36\implies 16x^2=36 \\\\\\ x^2=\cfrac{36}{16}\implies x^2 = \cfrac{9}{4}\implies x=\sqrt{\cfrac{9}{4}}\implies x=\cfrac{\sqrt{9}}{\sqrt{4}}\implies x = \cfrac{3}{2}~~\textit{seconds}$

now, we know the 2nd time around it hit the ground, h(x) = 0, but it took less time, it took 0.5 or 1/2 second less, well, the first time it took 3/2, if we subtract 1/2 from it, we get 3/2 - 1/2 = 2/2 = 1, so it took only 1 second this time then, meaning x = 1.

$\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(x) = -16x^2+v_ox+h_o \quad \begin{cases} v_o=\textit{initial velocity}&0\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&0\\ \qquad \textit{at "t" seconds}\\ x=\textit{seconds}&1 \end{cases} \\\\\\ 0=-16(1)^2+0x+h_o\implies 0=-16+h_o\implies 16=h_o \\\\[-0.35em] ~\dotfill\\\\ ~\hfill h(x) = -16x^2+16~\hfill$

quick info:

in case you're wondering what's that pesky -16x² doing there, is gravity's pull in ft/s².

4 0

3 years ago