Answer:
12
Step-by-step explanation:
1.6 I think I’m sorry if it’s incorrect
Answer:
<h2><em><u>1st</u></em><em><u> </u></em><em><u>option</u></em><em><u> </u></em></h2>
Step-by-step explanation:
Place the compass at one end of line segment, and taking it more than 1/2 of MN is its 1st step
<span>We use the binomial distribution, which states that:
Probability(r out of n) = (nCr) (p)^r (q)^(n-r)
In this case, n = 10 students, p = 39% = 0.39, and q = 1 - 0.39 = 0.61
a) For r = 2: Probability(2/10) = (10C2) (0.39)^2 (0.61)^(10-2) = 45(0.39)^2 (0.61)^8 = 0.1312, which is the probability for exactly 2.
b) We can first find the probability for r = 0 and r = 1, then subtract that from 1 to determine the probability of at least 2.
For r = 0: </span><span>Probability(0/10) = (10C0) (0.39)^0 (0.61)^(10-0) = (1)(1)(0.61)^10 = 0.0071
</span><span>For r = 1: Probability(1/10) = (10C1) (0.39)^1 (0.61)^(10-1) = (10)(0.39)(0.61)^9 = 0.0456</span>
Then P(0/10) + P(1/10) = 0.0527, so P(at least 2/10) = 1 - 0.0527 = 0.9473.
(c)
We have P(2/10) = 0.1312, and we can calculate for the rest similarly:
For r = 3: <span>Probability(3/10) = (10C3) (0.39)^3 (0.61)^(10-3) = 0.2237
</span><span>For r = 4: <span>Probability(4/10) = (10C4) (0.39)^4 (0.61)^(10-4) = 0.2503
</span></span><span>For r = 5: <span>Probability(5/10) = (10C5) (0.39)^5 (0.61)^(10-5) = 0.1920
Therefore the sum of P(2) up to P(5) is 0.7972, so this is the probability of having between 2 to 5 inclusive.
</span></span>
Answer:
4a+20
Step-by-step explanation:
