The probability we consider factors that decide what engine is chosen and the odds of choosing an engine size greater than 3.9 L
<h3>
Probability of
choosing an engine size</h3>
An engine size greater than 3.9 L
Generally, This can be defined as the possibility of an event occurring with respect to other possible outcomes.
Probability looks at calculating the possibility of a given event's occurrence.
To determine the possibility of selecting an engine size greater than 3.9 L, we consider factors than decide what engine is chosen and the odds of choosing an engine size greater than 3.9 L
Hence, the probability we consider factors that decide what engine is chosen and the odds of choosing an engine size greater than 3.9 L
For more information on probability
brainly.com/question/795909
Răspuns:
125%
Explicație pas cu pas:
a și B sunt invers proporționale cu numerele 12 și 15
Proporționalitate inversă:
a = k / y
Prin urmare ;
a = k / 12 - - - (1)
b = k / 15 - - - (2)
Procent care reprezintă a de la b
a = 100%
b = x
k / 12 = 1
k / 15 = x
Înmulțirea încrucișată:
x * k / 12 = k / 15
x = k / 12 ÷ k / 15
x = k / 12 * 15 / k
x = 15/12
x = 1,25
1,25 * 100% = 125%
Answer:
It would be 23.94 pieces
Step-by-step explanation:
since he has one piece that is 84ft and wants 34ft pieces, you would divide 34 by 84 and get 23.94
Answer: 0.03855
Step-by-step explanation:
Given :A population of skiers has a distribution of weights with mean 190 pounds and standard deviation 40 pounds.
Its maximum safe load is 10000 pounds.
Let X denotes the weight of 50 people.
As per given ,
Population mean weight of 50 people = 
Standard deviation of 50 people 
Then , the probability its maximum safe load will be exceeded =
![P(X>10000)=P(\dfrac{X-\mu}{\sigma}>\dfrac{10000-9500}{282.84})\\\\=P(z>1.7671-8)\\\\=1-P(z\leq1.7678)\ \ \ \ [\because\ P(Z>z)=P(Z\leq z)]\\\\=1-0.96145\ \ \ [\text{ By p-value of table}]\\\\=0.03855](https://tex.z-dn.net/?f=P%28X%3E10000%29%3DP%28%5Cdfrac%7BX-%5Cmu%7D%7B%5Csigma%7D%3E%5Cdfrac%7B10000-9500%7D%7B282.84%7D%29%5C%5C%5C%5C%3DP%28z%3E1.7671-8%29%5C%5C%5C%5C%3D1-P%28z%5Cleq1.7678%29%5C%20%5C%20%5C%20%5C%20%5B%5Cbecause%5C%20P%28Z%3Ez%29%3DP%28Z%5Cleq%20z%29%5D%5C%5C%5C%5C%3D1-0.96145%5C%20%5C%20%5C%20%5B%5Ctext%7B%20By%20p-value%20of%20table%7D%5D%5C%5C%5C%5C%3D0.03855)
Thus , the probability its maximum safe load will be exceeded = 0.03855
