The answer is 93.6
156x .6 = 93.6
Answer:
the probability that the sample mean is greater than 2.1 but less than 2.4 is 0.2444
Step-by-step explanation:
Given the data in the question;
x f(x) xp(x) x²p(x)
1 1/3 0.33333 0.33333
2 1/3 0.66667 1.33333
3 1/3 1.00000 3.0000
∑ 2.0000 4.6667
∑(xp(x)) = 2
∑(x²p(x)) = 4.6667
Variance σ² = ∑(x²) - ∑(x)² = 4.6667 - (2)² = 4.6667 - 4 = 0.6667
standard deviation σ = √variance = √0.6667 = 0.8165
Now since, n = 33 which is greater than 30, we can use normal approximation
for normal distribution z score ( x-μ)/σ
mean μ = 2
standard deviation = 0.817
sample size n = 33
standard of error σₓ = σ/√n = 0.817/√33 = 0.1422
so probability will be;
p( 2.1 < X < 2.4 ) = p(( 2.1-2)/0.1422) < x"-μ/σₓ < p(( 2.4-2)/0.1422)
= 0.70 < Z < 2.81
= 1 - ( 0.703 < Z < 2.812 )
FROM Z-SC0RE TABLE
= 1 - ( 0.25804 + 0.49752 )
= 1 - 0.75556
p( 2.1 < X < 2.4 ) = 0.2444
Therefore, the probability that the sample mean is greater than 2.1 but less than 2.4 is 0.2444
By applying algebraic handling on the two equations, we find the following three <em>solution</em> pairs: x₁ ≈ 5.693 ,y₁ ≈ 10.693; x₂ ≈ 1.430, y₂ ≈ 6.430; x₃ ≈ - 0.737, y₃ ≈ 4.263.
<h3>How to solve a system of equations</h3>
In this question we have a system formed by a <em>linear</em> equation and a <em>non-linear</em> equation, both with no <em>trascendent</em> elements and whose solution can be found easily by algebraic handling:
x - y = 5 (1)
x² · y = 5 · x + 6 (2)
By (1):
y = x + 5
By substituting on (2):
x² · (x + 5) = 5 · x + 6
x³ + 5 · x² - 5 · x - 6 = 0
(x + 5.693) · (x - 1.430) · (x + 0.737) = 0
There are three solutions: x₁ ≈ 5.693, x₂ ≈ 1.430, x₃ ≈ - 0.737
And the y-values are found by evaluating on (1):
y = x + 5
x₁ ≈ 5.693
y₁ ≈ 10.693
x₂ ≈ 1.430
y₂ ≈ 6.430
x₃ ≈ - 0.737
y₃ ≈ 4.263
By applying algebraic handling on the two equations, we find the following three <em>solution</em> pairs: x₁ ≈ 5.693 ,y₁ ≈ 10.693; x₂ ≈ 1.430, y₂ ≈ 6.430; x₃ ≈ - 0.737, y₃ ≈ 4.263.
To learn more on nonlinear equations: brainly.com/question/20242917
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The expression 7•3x is not equal to the expression 21x. You might think that the two expressions are the same since 7x3 is 21 but the first expression is a dot product. This kind of expression includes the magnitude and direction of the vector in an expression, for example, <span>7•3x. The second expression, 21x, expresses only the magnitude and does not include the direction.</span>
Answer: The numerator and denominator degrees of freedom (respectively) for the critical value of F are <u>4</u> and<u> 118 .</u>
Step-by-step explanation:
We know that , for critical value of F, degrees of freedom for numerator = k-1
and for denominator = n-k, where n= Total observations and k = number of independent variables.
Here, Numbers of independent variables(k) = 5
Total observations (n)= 123
So, Degrees of freedom for numerator = 5-1=4
Degrees of freedom for denominator =123-5= 118
Hence, the numerator and denominator degrees of freedom (respectively) for the critical value of F are <u>4</u> and<u> 118 .</u>
