0.00
*First column to your left - ones column (1)
*Column just after decimal place, in the middle - tenths column (1/10)
*Column to the far right - hundredths column (1/100)
The only numbers that can go in these columns are 0-9.
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1.54 + 2.37
= (1 + 5/10 + 4/100) + (2 + 3/10 + 7/100)
= 1 + 5/10 + 4/100 + 2 + 3/10 + 7/100
= 3 + 8/10 + 11/100
= 3 + 8/10 + (10/100 + 1/100)
= 3 + 8/10 + (1/10 + 1/100)
= 3 + 8/10 + 1/10 + 1/100
= 3 + 9/10 + 1/100
= 3.91
Add 37 + 9 = 46
If thats not your answer theres more.
46 divided by 5 = 9.2
<span><span>1.
If we get an odd number first (probability 5/10 = 1/2), then the only way to get a number greater than 9 is the 10 (probability 1/9, since only 9 papers remain). This gives 1/2 * 1/9 = 1/18.</span></span><span><span>
</span></span><span><span>
</span></span><span><span>2.
We would expect that P(2 white cards) = P(1 white card)*P(2nd white card given 1 white card).</span></span><span><span /></span><span><span>This gives the equation 14/95 = 2/5 * x<span>
</span></span></span><span><span><span>Solving for x gives x = 7/19.</span></span></span><span><span>
</span></span><span><span>
</span></span><span><span>3. There are a total of 4 favorable results: 1, 4, 5, 6. Out of the 6 possibilities, this is a probability of 4/6 = 2/3.</span></span>
Answer:
Step-by-step explanation:
- Use of low <em>of sines</em>
- <em>Consider triangle CEN</em>
- <em>Bearing of angle counted from zero to the vector of flight</em>
<u>We have </u>
- ∠CNE = complementary with 44° ⇒ m∠CNE = 90° - 44° = 46°
- NE = b = 545 km
- EC = a = 680 km
- ∠ECN = x
<u>Find the measure of angle x:</u>
- sin 46° / a = sin x / b
- sin 46° / 680 = sin x / 545
- sin x = 545 sin 46 deg / 680
- x = arcsin (545 sin 46 deg / 680)
- x = 35° (rounded)
<u>Bearing of the flight from Elgin to Canton is:</u>
- 270° - 35° = 235° as the vector is in the third quadrant
We have to assume that the intention is that S represents the set of possible outcomes.
A. TRUE — any of the faces may appear
B. false — 0 is not a possible outcome and is not in set S.
C. TRUE — {5, 6} is a subset of S
D. TRUE — A = S \{5}
E. false — even numbers are {2, 4, 6}. The complement of that includes 5, which A does not.
