There's 2 easy ways to do it (more than 2, but I find these easy):
First of all, divide the price by 100 to find what 1% is.
Now, you can either multiply by thirty to find how big the discount is and take that away OR you can multiply by 70, seeing as 70% is left once the 30% discount is in effect.
Answer:
a = 7
Step-by-step explanation:
Using Pythagoras' identity in the right triangle,
The square on the hypotenuse is equal to the sum of the squares on the other 2 sides, that is
a² + b² = c² , substitute values
a² + 2.4² = 7.4², that is
a² + 5.76 = 54.76 ( subtract 5.76 from both sides )
a² = 49 ( take the square root of both sides )
a =
= 7
The answer is d, (-2.5,0.75)
Solution for 0.5 is what percent of 3:
0.5:3*100 =
(0.5*100):3 =
50:3 = 16.666666666667
Now we have: 0.5 is what percent of 3 = 16.666666666667
Question: 0.5 is what percent of 3?
Percentage solution with steps:
Step 1: We make the assumption that 3 is 100% since it is our output value.
Step 2: We next represent the value we seek with $x$x.
Step 3: From step 1, it follows that $100\%=3$100%=3.
Step 4: In the same vein, $x\%=0.5$x%=0.5.
Step 5: This gives us a pair of simple equations:
$100\%=3(1)$100%=3(1).
$x\%=0.5(2)$x%=0.5(2).
Step 6: By simply dividing equation 1 by equation 2 and taking note of the fact that both the LHS
(left hand side) of both equations have the same unit (%); we have
$\frac{100\%}{x\%}=\frac{3}{0.5}$
100%
x%=
3
0.5
Step 7: Taking the inverse (or reciprocal) of both sides yields
$\frac{x\%}{100\%}=\frac{0.5}{3}$
x%
100%=
0.5
3
$\Rightarrow x=16.666666666667\%$⇒x=16.666666666667%
Therefore, $0.5$0.5 is $16.666666666667\%$16.666666666667% of $3$3.
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.