Answer:
0.7486 = 74.86% observations would be less than 5.79
Step-by-step explanation:
I suppose there was a small typing mistake, so i am going to use the distribution as N (5.43,0.54)
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.
The general format of the normal distribution is:
N(mean, standard deviation)
Which means that:
What proportion of observations would be less than 5.79?
This is the pvalue of Z when X = 5.79. So
has a pvalue of 0.7486
0.7486 = 74.86% observations would be less than 5.79
Answer: see proof below
<u>Step-by-step explanation:</u>
Use the Double Angle Identity: sin 2Ф = 2sinФ · cosФ
Use the Sum/Difference Identities:
sin(α + β) = sinα · cosβ + cosα · sinβ
cos(α - β) = cosα · cosβ + sinα · sinβ
Use the Unit circle to evaluate: sin45 = cos45 = √2/2
Use the Double Angle Identities: sin2Ф = 2sinФ · cosФ
Use the Pythagorean Identity: cos²Ф + sin²Ф = 1
<u />
<u>Proof LHS → RHS</u>
LHS: 2sin(45 + 2A) · cos(45 - 2A)
Sum/Difference: 2 (sin45·cos2A + cos45·sin2A) (cos45·cos2A + sin45·sin2A)
Unit Circle: 2[(√2/2)cos2A + (√2/2)sin2A][(√2/2)cos2A +(√2/2)·sin2A)]
Expand: 2[(1/2)cos²2A + cos2A·sin2A + (1/2)sin²2A]
Distribute: cos²2A + 2cos2A·sin2A + sin²2A
Pythagorean Identity: 1 + 2cos2A·sin2A
Double Angle: 1 + sin4A
LHS = RHS: 1 + sin4A = 1 + sin4A 
5x-13=15+7x
-13=15+2x
-28=2x
-14=x
Subtract 4 from both sides
y−4≤−2x
Divide both sides by −2
- y-4/2 ≥ x
Switch sides
<span>x ≤ − y−4/2<span><span><span><span><span></span></span></span>
HOPE THIS HELPS!!
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