Complete Question
A set of magical wand prices are normally distributed with a mean of 50 dollars and a standard deviation of 4 dollars. A blackthorn wand has a price of 45.20. What proportion of wand prices are lower than the price of the blackthorn wand? You may round your answer to four decimal places
Answer:
0.1151
Step-by-step explanation:
We solve using z score formula
z = (x-μ)/σ, where
x is the raw score = $45.20
μ is the population mean = $50
σ is the population standard deviation = $4
We are solving for x < 45.20
Hence:
z = 45.20 - 50/4
z = -1.2
Probability value from Z-Table:
P(x<45.20) = 0.11507
Approximately to 4 decimal places = 0.1151
Therefore, the proportion of wand prices that are lower than the price of the blackthorn wand is 0.1151
Step-by-step explanation:
AB = 3
by similaritiy criterion
CE/ CB= ED/AB
4/6= 2/AB
AB = 3
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R1 ÷ 2 = ![\left[\begin{array}{ccc}1&3&1.5\\-5&1&4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%263%261.5%5C%5C-5%261%264%5Cend%7Barray%7D%5Cright%5D)
R2 ÷ -5 = ![\left[\begin{array}{ccc}1&3&1.5\\1&-0.2&-0.8\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%263%261.5%5C%5C1%26-0.2%26-0.8%5Cend%7Barray%7D%5Cright%5D)
R2: R1 - R2 = ![\left[\begin{array}{ccc}1&3&1.5\\0&3.2&2.3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%263%261.5%5C%5C0%263.2%262.3%5Cend%7Barray%7D%5Cright%5D)
R2 ÷ 3.2 = ![\left[\begin{array}{ccc}1&3&1.5\\0&1&0.71875\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%263%261.5%5C%5C0%261%260.71875%5Cend%7Barray%7D%5Cright%5D)
R1: R1 - 3R2 = ![\left[\begin{array}{ccc}1&0&0.65625\\0&1&0.71875\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%260.65625%5C%5C0%261%260.71875%5Cend%7Barray%7D%5Cright%5D)
Answer: x = 0.65625, y = 0.71875
Answer:
the answer to your question is A
Answer:
3 + 1(3) +3 (the dollar amounts)
3+3+3
9
0.5 + 0.1 +0.1 +0.1 +0.5 (the cents)
1.3
9 +1.3 = 10.30 worth of groceries using front end estimation
