Find three consecutive integers whose sum is six times the greater integer
1 answer:
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Answer:
0.1994 is the required probability.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 166 pounds
Standard Deviation, σ = 5.3 pounds
Sample size, n = 20
We are given that the distribution of weights is a bell shaped distribution that is a normal distribution.
Formula:
Standard error due to sampling =
P(sample of 20 boxers is more than 167 pounds)
Calculation the value from standard normal z table, we have,
0.1994 is the probability that the mean weight of a random sample of 20 boxers is more than 167 pounds
Assuming a diagram similar to the one I've attached, ∠<em>YVZ</em> is a vertical angle to ∠<em>WVX</em>, which means they have an equal measure. Additionally, ∠<em>WVZ</em> and ∠<em>WVX</em> form a linear pair, which means they are supplementary (sum to 180°). That means we start out with the equation
We combine our like terms (the <em>x</em>'s get combined, then the constants get combined) and have:
Cancel the 9 first by subtraction:
Cancel the 19 by division:
Since we know that our angle we're looking for, ∠<em>YVZ</em>, is the same measure as ∠<em>WVX</em>, we substitute 9 in for <em>x</em>:
8(9)+28=72+28=100°
We combine our like terms (the <em>x</em>'s get combined, then the constants get combined) and have:
Cancel the 9 first by subtraction:
Cancel the 19 by division:
Since we know that our angle we're looking for, ∠<em>YVZ</em>, is the same measure as ∠<em>WVX</em>, we substitute 9 in for <em>x</em>:
8(9)+28=72+28=100°