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Sidana [21]
3 years ago
6

A candy bar takes an average of 3.4 hours to move through an assembly line. If the standard deviation is 0.5 hours, what is the

probability that the candy bar will take between 3 and 4 hours to move through the line?. A:.2119 B:.2295 C:.6731 D:.3270 E:.8849. 2. the average noise level in a diner is 30 decibels with a standard deviation of 6 decibels. 99% of the time, the noise level is below what value?. A: 16.04 B: 30.00 C: 36.00 D: 43.96 E: 48.00.
Mathematics
1 answer:
stepladder [879]3 years ago
7 0
To answer the first problem, we determine first the equivalent z score and get the corresponding probability. The z score at 3 hours is (3-3.4)/0.5 = 0.8 while that of 4 hrs is <span>(4-3.4)/0.5 = 1.2. z score of 0.8 is equal to a probability of 0.2119 and that at 1.2 is 0.8849. The difference is 0.673. Hence answer is C. 
For problem 2, the z score at 99% is 2.33 hence, the equation becomes 2.33 = (x-30)/6 where x is the noise level threshold. Answer is D. 43.96 decibels.</span>
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3 years ago
Can somebody explain why 8/17 is correct. Will make brainliest.
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OC bisects AOB, OD bisects AOC, OE bisects AOD, OF bisects AOE, OG bisects BOF. if BOF =120°, find DOE
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Answer:

∠DOE = 16°

Step-by-step explanation:

The given parameters are;

∠BOF = 120°

∠AOB = 2×∠AOC                        {}      Given

∠AOC = 2×∠AOD                     {}         Given

∠AOD = 2×∠AOE      {}                         Given

∠AOE = 2×∠AOF {}                               Given

Therefore;

∠AOB = 16×∠AOF                 {}              Angle addition postulate

∠BOF = ∠AOB - ∠AOF = 16×∠AOF  - ∠AOF = 15×∠AOF {} Transitive property

15×∠AOF = 120°  

∠AOF = 120°/15 = 8°

Given that OE bisects ∠AOD, we have;

∠AOE ≅ ∠DOE                      {}                    Angles bisected by a line

From;

∠AOE = 2×∠AOF, we have;  {}                               Given

Therefore;

∠AOE = ∠DOE = 2×∠AOF = 2×8° = 16°

∠DOE = 16°.

8 0
3 years ago
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