Answer:
11
Step-by-step explanation:
11 + 8 + 14 + v + 8 + 14
================== = 11 Combine Terms
6
(55 + v) / 6 = 11 Multiply by 6
55 + v = 11 * 6 Combine the right
55 + v = 66 Subtract 55
55-55+v = 66- 55
v = 11
Answer:
Step 1: If the radicals have the same index, multiply terms the outside the radical with terms outside the radical and terms inside the radical with terms inside the radical.
Step 2: Simplify the radicals.
Step 3: Multiply the terms outside the radical
Step-by-step explanation:
Hope this helps!
Answer:
The most that a bag can weigh and not need to be repackaged is 15.355 ounces.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation , the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
Bags in the upper 4% are too heavy and must be repackaged. What is the most that a bag can weigh and not need to be repackaged?
Bags in the upper 4% have a pvalue of 1-0.04 = 0.96. So the most that a bag can weight and not need to be repackaged is the value of X when Z has a pvalue of 0.9599. So this is X when
The most that a bag can weigh and not need to be repackaged is 15.355 ounces.
Answer:
See answers below
Step-by-step explanation:
The Bisector Angle Theorem basically states that if a bisector cuts the middle of an angle, the two angles that it creates will be equal to each other.
Therefore:
#7 -> x=38
#8 -> x=144
#9 -> 2x=84 -> x=42
#10 -> x+3=30 -> x=27