The answer is b, because u need to use m= y2-y1/x2-x1
-2(b-1)=2(1+7b)
First, distribute; -2b+2=2+14b
Get the variables on one side; -2b+2=2+14b
-14b -14b
Get the constants on one side; -16b+2=2
-2 -2
Divide to find “b”; -16b/-16=0/-16
b=0
Using the probability concept, we have that:
a) It would not be unusual to observe one component fail, since the probability that one component fails is greater than 0.05.
b) It would be unusual to observe two components fail, since the probability that two components fail is less than 0.05.
<h3>What is a probability?</h3>
A probability is given by the <u>number of desired outcomes divided by the number of total outcomes</u>. If a probability is less than 0.05, the event is considered unusual.
In this problem, the probabilities are given as follows:
- 0.21 probability that one component fails, hence not unusual.
- (0.21)² = 0.0441 probability that two components fail, hence unusual.
More can be learned about probabilities at brainly.com/question/14398287
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Answer:
z0.07/2=1.81
Step-by-step explanation:
thus:z0.035=1.81
z0.07/2=1.81
Answer:
Yes this compound could be shown to be butane, specifically, it can be shown to be 9 molecules of butane. Check Explanation for more.
Step-by-step explanation:
Butane contains 10 hydrogen atoms for every 4 carbon atoms. It's molecular formula is C₄H₁₀
The hydrocarbon compound presented has 36 carbon atoms and 90 hyfrogen atoms
If the ratio of each of these atoms present is also in the ratio 4:10 like C:H, then, it can be inferred that the compound is indeed Butane
C | H
36 | 90
dividing through by 9, we have
C | H
4 | 10
which is the exact ratio of Carbon to Hydrogen in butane.
Hence, the unknown compound is most likely 9 molecules of butane
9C₄H₁₀ = 36 carbon atoms and 90 hydrogen atoms.
Hope this Helps!!!
