Answer:
1/2 or 0.5
Step-by-step explanation:
Within 1 to 6,
Even numbers are 2, 4, 6
While odd numbers are 1, 3, 5
Assuming cards are mixed up properly and the probabilities of drawing a card with any number from 1 to 6 is constant,
Probability of drawing even number cards = number of units of even number cards ÷ total number of units of cards from 1 to 6 aka total number of cards
= 3 / 6
= 1 / 2
Teri had 7/9 of a sandwich, you need to find a common denominator for 2/3. Multiply 2/3 by 3 which is 6/9. (6/9 is how much they ate) Then you subtract 7/9 by 6/9 (7/9-6/9=1/9)
Answer:1/9 of a yard
I hope this helped.. not much of a teacher hahah :)
Answer:
the question?
Step-by-step explanation:
Answer:
x-2=x x=0
No Solution
x+1=1-x
Cancel
1 on both sides
Subtract
−x from both sides
x+x=0
Simplify x
+x to 2x
2x=0
x=0
Check Answer
x+1=1-x
0+1=1-0
1=1-0
1=1
Answer:
So then the correct answer would be:
B) .9996
Step-by-step explanation:
The exact way to solve this problem is using the binomial distribution, assuming that our random variable of interest is "number of students living in apartments" represented by X and 
And we want this probability:
![P(200 \leq X \leq 400) [tex]In order to find this probability we can use the foloowing excel code:"=BINOM.DIST(400,600,0.4,TRUE)-BINOM.DIST(199,600,0.4,TRUE)"And we got:[tex] P(200 \leq X \leq 400) =0.999675[tex]But for this case the problem says that we need to approximate, so then we can use the normal approximation to the normal distribution. We need to check the conditions in order to use the normal approximation. [tex]np=600*0.4=240\geq 10](https://tex.z-dn.net/?f=%20P%28200%20%5Cleq%20X%20%5Cleq%20400%29%20%5Btex%5D%3C%2Fp%3E%3Cp%3EIn%20order%20to%20find%20this%20probability%20we%20can%20use%20the%20foloowing%20excel%20code%3A%3C%2Fp%3E%3Cp%3E%22%3DBINOM.DIST%28400%2C600%2C0.4%2CTRUE%29-BINOM.DIST%28199%2C600%2C0.4%2CTRUE%29%22%3C%2Fp%3E%3Cp%3EAnd%20we%20got%3A%3C%2Fp%3E%3Cp%3E%5Btex%5D%20P%28200%20%5Cleq%20X%20%5Cleq%20400%29%20%3D0.999675%5Btex%5D%3C%2Fp%3E%3Cp%3EBut%20for%20this%20case%20the%20problem%20says%20that%20we%20need%20to%20approximate%2C%20so%20then%20we%20can%20use%20the%20normal%20approximation%20to%20the%20normal%20distribution.%20%3C%2Fp%3E%3Cp%3EWe%20need%20to%20check%20the%20conditions%20in%20order%20to%20use%20the%20normal%20approximation.%0A%3C%2Fp%3E%3Cp%3E%5Btex%5Dnp%3D600%2A0.4%3D240%5Cgeq%2010)
So we see that we satisfy the conditions and then we can apply the approximation.
If we appply the approximation the new mean and standard deviation are:
And then 
And we are interested on the following probability:
So then the correct answer would be:
B) .9996
