Answer:
The answer is $26.7
Step-by-step explanation:yes
![\frac{1}{2.54} =\frac{24}{x}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2.54%7D%20%3D%5Cfrac%7B24%7D%7Bx%7D)
Multiply both sides by 2.54.
![1=\frac{24}{x} (2.54)](https://tex.z-dn.net/?f=1%3D%5Cfrac%7B24%7D%7Bx%7D%20%282.54%29)
![= \frac{60.96}{x}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B60.96%7D%7Bx%7D)
Multiply both sides by x.
x = 60.96
Hence, Aaron needs 60.96 centimeters of wire.
The equation which is equivalent to 60% of 25 is x = 0.6 * 25
An equation is used to show the relationship between variables, numbers.
Let x represent the value of 60% of 25. Hence:
x = 60% of 25
x = 0.6 * 25
Hence the equation which is equivalent to 60% of 25 is x = 0.6 * 25
x/25 = 60/100
6/10 = x/25
Answer:
a) C. No, a carton can have a puncture and a smashed corner.
b) The probability that a carton has a puncture or a smashed corner is P(X ∪ Y) = 0.104.
Step-by-step explanation:
To be mutually exclusive, the probability of the two events happening at the same time should be 0. But the probability that a carton has a puncture and has a smashed corner is 0.004 and not 0.
Then, we can conclude the events "selecting a carton with a puncture" and "selecting a carton with a smashed corner" are not mutually exclusive.
The answer is "C. No, a carton can have a puncture and a smashed corner."
We can calculate the probability that the carton has a puncture <em>or </em>has a smashed comer simply by adding the probability of each event:
![P(X\cup Y)=P(X)+P(Y)=0.1+0.04=0.104](https://tex.z-dn.net/?f=P%28X%5Ccup%20Y%29%3DP%28X%29%2BP%28Y%29%3D0.1%2B0.04%3D0.104)
P(X ∪ Y): probability that a carton has a puncture or a smashed corner.
$210.67 is the answer. Rounded, of course, from 210.666666667