Answer:
Step-by-step explanation:
Given
Required
Factor
Factor out 
<em>The expression cannot be further factored</em>
If scores on an exam follow an approximately normal distribution with a mean of 76.4 and a standard deviation of 6.1 points, then the minimum score you would need to be in the top 2% is equal to 88.929.
A problem of this type in mathematics can be characterized as a normal distribution problem. We can use the z-score to solve it by using the formula;
Z = x - μ / σ
In this formula the standard score is represented by Z, the observed value is represented by x, the mean is represented by μ, and the standard deviation is represented by σ.
The p-value can be used to determine the z-score with the help of a standard table.
As we have to find the minimum score to be in the top 2%, p-value = 0.02
The z-score that is found to correspond with this p-value of 0.02 in the standard table is 2.054
Therefore,
2.054 = x - 76.4 ÷ 6.1
2.054 × 6.1 = x - 76.4
12.529 = x - 76.4
12.529 + 76.4 = x
x = 88.929
Hence 88.929 is calculated to be the lowest score required to be in the top 2%.
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Answer:
<h2>The year is 2010.</h2>
Step-by-step explanation:
The population of wolves is given by P(t) = 80(0.98)t.
The study has began in the year of 2000 and will end in 2020.
The difference between 2000 and 2009 is 9 years and the difference between 2009 and 2020 is 11 years.
Hence, the maximum value of t is 11.
The domain is 
.
The population will be 80, that is P(t) = 80.
Hence, 0.98t = 1, or, t = 
≅1.
In (2009 + 1) = 2010, the population of wolves will be 80.
Answer:
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Step-by-step explanation:
Step-by-step explanation:
let necklace : x
let earrings : y
x+y=144 -1st eqn
6x+5y=825 -2nd eqn
y=144-X -3rd eqn
substitute eqn into 2nd eqn
6x+5(144-x)= 825
6x +720-5X =825 (expanded)
X = 825-720
X =105
substitute "X=105" into 1st eqn
(105)+y=144
y= 144-105
y=39
1 necklace = 105$
1 pair of earrings = 39$
(to check back: substitute the value into the original eqn)
