The answer is B.
If a coin is flipped 14 times and lands on heads 10 times the fraction would equal to 10/14.
10/14 simplifies to 5/7.
Answer:
The minimum weight for a passenger who outweighs at least 90% of the other passengers is 203.16 pounds
Step-by-step explanation:
Problems of normally distributed samples are 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:

What is the minimum weight for a passenger who outweighs at least 90% of the other passengers?
90th percentile
The 90th percentile is X when Z has a pvalue of 0.9. So it is X when Z = 1.28. So




The minimum weight for a passenger who outweighs at least 90% of the other passengers is 203.16 pounds
Answer:
A) -2 - i√3 , -2 + i√3
Step-by-step explanation:
Solve using quadratic formula
x² + 4x + 7 = 0
The Almighty Formula
= -b ± √b² - 4ac/2a
Where ax + bx² + c = 0
From the above question
a = 1, b = 4, c= 7
Hence,
-4 ± √4² - 4 × 1 × 7/2 × 1
-4 ± √16 - 28/2
=( -4 ± √-12)/2
Since
b² - 4ac < 0
We have two complex roots
Simplifying
( -4 ± √-12)/2
= -4/2 ± √-12/2
= -2 ± 2√3i/2
= -2 ± √3i
Therefore,
-2 - √3i , -2 + √3i
or
-2 - i√3 , -2 + i√3
Option A , is the correct answer