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Answer:
The probability that the mean daily precipitation will be 0.11 inches or less for a random sample of November days
P(X≤ 0.11) = 0.4404
Step-by-step explanation:
<u><em>Step(i):-</em></u>
Given that the mean of the Population = 0.10 inches
Given that the standard deviation of the population = 0.07inches
Let 'X' be a random variable in a normal distribution

<u><em>Step(ii):-</em></u>
The probability that the mean daily precipitation will be 0.11 inches or less for a random sample of November days
P(X≤ 0.11) = P(Z≤0.1428)
= 1-P(Z≥0.1428)
= 1 - ( 0.5 +A(0.1428)
= 0.5 - A(0.1428)
= 0.5 -0.0596
= 0.4404
<u><em>Final answer:-</em></u>
The probability that the mean daily precipitation will be 0.11 inches or less for a random sample of November days
P(X≤ 0.11) = 0.4404
Answer:
√13 ≈ 3.6056
Step-by-step explanation:
The Law of Cosines can be used to figure this. If the third side is "c", then it tells you ...
c² = a² + b² - 2ab·cos(C)
c² = 3² + 4² -2(3)(4)(cos(60°)) = 9 + 16 - 24(1/2) = 13
c = √13 ≈ 3.6056
The length of the third side is √13, about 3.6056.
Answer:
i dont see a table
Step-by-step explanation: