Yes...the person's hypothesis could be correct.
Answer:
The answer to your question is:
x = 1
y = 1
z = 0
Step-by-step explanation:
-2x + 2y + 3z = 0 (1)
-2x - y + z = -3 (2)
2x + 3y + 3z = 5 (3)
Solve (1) and (2)
Multiply 2 by 2
-2x + 2y + 3z = 0
-4x -2y + 2z = -6
-6x + 5 z = -6 (4)
Solve (2) and (3)
Multiply 2 by 3
-6x - 3y + 3z = -9
2x + 3y + 3z = 5
-4x + 6z = -4 (5)
Solve (4) and (5)
Multiply (4) by 2 and (5) by -3
-12x + 10 z = -12
12x - 18z = 12
-6z = 0
z = 0
Then
-4x + 6(0) = -4
-4x = -4
x = -4/-4
x = 1
Finally
-2(1) - y + (0) = -3
-2 - y = -3
-y = -3 + 2
y = 1
Answer:
Step-by-step explanation:
b
Answer:
$31.8
Step-by-step explanation:
Assuming the 6% tax is separate from the $30, we can use this equation:
Total = Cost + Tax (pretty self explanatory)
Method 1
Total = $30 + (6% * $30)
Total = $30 + $1.8
Total = $31.8
Method 2
Total = $30 * 1.06 (The 1 is the original cost, the .06 is the 6% tax)
Total = $31.8
Answer:
Probability that average height would be shorter than 63 inches = 0.30854 .
Step-by-step explanation:
We are given that the average height of 20-year-old American women is normally distributed with a mean of 64 inches and standard deviation of 4 inches.
Also, a random sample of 4 women from this population is taken and their average height is computed.
Let X bar = Average height
The z score probability distribution for average height is given by;
Z = 
~ N(0,1)
where, 
= population mean = 64 inches
= standard deviation = 4 inches
n = sample of women = 4
So, Probability that average height would be shorter than 63 inches is given by = P(X bar < 63 inches)
P(X bar < 63) = P( < 
) = P(Z < -0.5) = 1 - P(Z <= 0.5)
= 1 - 0.69146 = 0.30854
Hence, it is 30.85% likely that average height would be shorter than 63 inches.
