my mother is 12 years more than twice my age.after 8 years she will be 20 years less than three times my age...find my age and m
1 answer:
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Given :
Challenge given is to volunteer for about 1 1/2 hours per week .
Autum did at a rate of 6 3/4 hours per week .
To Find :
The comparison ( ratio ) her volunteers rate with the challenge rate .
Solution :
Simplifying challenge rate in in normal fraction , .
Also , simplifying her rate , .
Now , ratio of his work by challenge is :
Therefore , the ratio is 9/2 or 4 1/2 .
Hence , this is the required solution .
Answer:
(b)0.56
(c)0.38
Step-by-step explanation:
(a)
P(Ben Pass) =0.8
Therefore: P(Ben fails)=1-0.8 =0.2
P(Tom Pass) =0.7
Therefore: P(Tom fails)=1-0.7 =0.3
See attached for the completed tree diagram
(b)Probability that both will pass
P(both will pass)=P(Ben pass and Tom pass)
=P(Ben pass) X P(Tom pass)
=0.8 X 0.7
=0.56
(c)The probability that only one of them will pass
Since either Tom or Ben can pass, we have:
P(only one of them will pass)
=P(Ben pass and Tom fails OR Ben Fails and Tom Pass)
=P(Ben pass and Tom fails)+P(Ben Fails and Tom Pass)
=(0.8 X 0.3) + (0.2 X 0.7)
=0.24 + 0.14
=0.38
Answer:
The designed life should be of 21,840 vehicle miles.
Step-by-step explanation:
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean and standard deviation
, the z-score of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value 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. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Mean of 35,000 vehicle miles and a standard deviation of 7,000 vehicle miles.
This means that
Find its designed life if a .97 reliability is desired.
The designed life should be the 100 - 97 = 3rd percentile(we want only 3% of the vehicles to fail within this time), which is X when Z has a p-value of 0.03, so X when Z = -1.88.
The designed life should be of 21,840 vehicle miles.