Answer:

Step-by-step explanation:
In order to find the side mentioned (IJ), we need to use SOH CAH TOA.
SOH CAH TOA is an acronym to help us remember what sin, cos, and tan mean. It stands for:
Sin = Opposite / Hypotenuse
Cosine = Adjacent / Hypotenuse
Tan = Opposite / Adjacent
Since we know the measure of angle K (42) and we know one of the sides, we can use this to find the missing length.
Since the side given to us is the hypotenuse, and we're looking for the side opposite of the angle (IJ), the only possible one to use would be SIN as it includes Opposite and Hypotenuse.
Our equation is now this: 
Let's now solve for x.
Therefore, the length of IJ will be around 2.01.
Hope this helped!
- Let x be the percentage of p.
- Therefore,

<u>Answer:</u>
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Hope you could get an idea from here.
Doubt clarification - use comment section.
Answer:
1.92%
Step-by-step explanation:
The probability for first case, picking a queen out of deck, will be:

as there will be 4 queens in a deck, one of each suit.
For the second pick, the probability of picking a diamond card, will be:

here the total will remain 52 as he has replaced the first card and not kept it aside and there will be 13 cards in diamond suit (including the three face cards).
Thus the net probability for both cases will be:

Thus total probability for the combined two cases will be 1.92%