Find the ratio that is equivalent to 10:5. OA) 15:5 OB) 5:10 OC) 2:1 OD 5:1
2 answers:
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Answer:
46.18% of the items will weigh between 6.4 and 8.9 ounces.
Step-by-step explanation:
We are given that the weight of items produced by a machine is normally distributed with a mean of 8 ounces and a standard deviation of 2 ounces.
<em>Let X = weight of items produced by a machine</em>
The z-score probability distribution for is given by;
Z = ~ N(0,1)
where, = mean weight = 8 ounces
= standard deviation = 2 ounces
The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.
So, percentage of items that will weigh between 6.4 and 8.9 ounces is given by = P(6.4 < X < 8.9) = P(X < 8.9 ounces) - P(X 6.4 ounces)
P(X < 8.9) = P( <
) = P(Z < 0.45) = 0.67364 {using z table}
P(X 70) = P(
) = P(Z
-0.80) = 1 - P(Z < 0.80)
= 1 - 0.78814 = 0.21186
<em>Now, in the z table the P(Z </em><em> x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 0.45 and x = 0.80 in the z table which has an area of </em>0.67364<em> and </em>0.78814<em> respectively.</em>
Therefore, P(6.4 < X < 8.9) = 0.67364 - 0.21186 = 0.4618 or 46.18%
<em>Hence, 46.18% of the items will weigh between 6.4 and 8.9 ounces.</em>
Check the picture below.
the segment LOJ is a diameter, and therefore the arcLCJ made by that diameter will have a central angle of 180°.
that simply means that the arcLC is 180 - 94, or 86°.
the ∡LJC is an inscribed angle that is intercepting the arcLC, and therefore, by the inscribed angle theorem, it'll be half of arcLC.
the segment LOJ is a diameter, and therefore the arcLCJ made by that diameter will have a central angle of 180°.
that simply means that the arcLC is 180 - 94, or 86°.
the ∡LJC is an inscribed angle that is intercepting the arcLC, and therefore, by the inscribed angle theorem, it'll be half of arcLC.