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3 years ago

I cant solve this i need help

Mathematics

1 answer:

tangare [24]3 years ago

6 0

Hey!

Here is the equation:

$\frac{j}{3} = 4.5$

Switch the equation around and it becomes:

$4.5 \times 3 = j$

The sign changes when moved to the other sign. At first it was (÷) and now it's (×).

Multiply 4.5 by 3 to find j

$4.5 \times 3 = 13.5$

<em>That means your answer is 13.5</em>

$\framebox{j = 13.5}$

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see the attached figure to better understand the problem

we know that

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3 years ago

Can you please help me out

Taya2010 [7]

The bag contains,

Red (R) marbles is 9, Green (G) marbles is 7 and Blue (B) marbles is 4,

Total marbles (possible outcome) is,

$\text{Total marbles = (R) + (G) +(B) = 9 + 7 + 4 = 20 marbles}$

Let P(R) represent the probablity of picking a red marble,

P(G) represent the probability of picking a green marble and,

P(B) represent the probability of picking a blue marble.

Probability , P, is,

$\text{Prob, P =}\frac{required\text{ outcome}}{possible\text{ outcome}}$ $\begin{gathered} P(R)=\frac{9}{20} \\ P(G)=\frac{7}{20} \\ P(B)=\frac{4}{20} \end{gathered}$

Probablity of drawing a Red marble (R) and then a blue marble (B) without being replaced,

That means once a marble is drawn, the total marbles (possible outcome) reduces as well,

$\begin{gathered} \text{Prob of a red marble P(R) =}\frac{9}{20} \\ \text{Prob of }a\text{ blue marble =}\frac{4}{19} \\ \text{After a marble is selected without replacement, marbles left is 19} \\ \text{Prob of red marble + prob of blue marble = P(R) + P(B) = }\frac{9}{20}+\frac{4}{19}=\frac{251}{380} \\ \text{Hence, the probability is }\frac{251}{380} \end{gathered}$

Hence, the best option is G.

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1 year ago