<span>Because G is the centroid of the triangle, therefore
AG = 2GF
That is,
5x + 4 = 2(3x - 1)
5x + 4 = 6x - 2
4 + 2 = 6x - 5x
6 = x
Therefore
AG = 5*6 + 4 = 34
GF = 3*6 - 1 = 17
A.F = AG + GF = 34 + 17 = 51
Answer: 51</span>
Answer: 120 cartons
Step-by-step explanation: Since, the grocery clerk displays 15 cartons at the base of the triangle and one at the top. Therefore, one carton is reduced at previous step and so on. Since, this series is in arithmetic progression. Therefore, 120 cartons are needed for the complete display.
I’ll explain part A and you can apply it to the other parts. The functions f(x), g(x), and m(x) are all equivalent to the given polynomials. So when it asks for f(x) + [g(x) - m(x)] you plug them in. The answer to Part A is 3x^2 +5x - z + [5x^2-3x+z - 2x^2 - 2x]. Combine the like terms (3x^2+5x^2-2x^2 = 6x^2 (because 3+5-2=6)) and so on until you’ve simplified it as much as possible!
<h3>
Answers:</h3><h3>
x = 12</h3><h3>
Angle B = 45 degrees</h3>
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Explanation:
Angles A and B are alternate exterior angles.
Due to the parallel lines, alternate exterior angles are congruent
A = B
5x-15 = 2x+21
5x-2x = 21+15
3x = 36
x = 36/3
x = 12
Use this x value to find the measure of angle B
angle B = 2x+21
angle B = 2*12+21
angle B = 24+21
angle B = 45 degrees
As a check,
angle A = 5x-15
angle A = 5*12+15
angle A = 60-15
angle A = 45 degrees
Since A and B are both 45 degrees, this confirms the correct x value
Answer:
Kindly check explanation
Step-by-step explanation:
H0 : μ = 5500
H1 : μ > 5500
The test statistic assume normal distribution :
Test statistic :
(Xbar - μ) ÷ s/sqrt(n)
(5625.1 - 5500) ÷ 226.1/sqrt(15) = 2.1429 = 2.143
Pvalue from test statistic :
The Pvalue obtained using the calculator at df = 15 - 1 = 14 is 0.025083
α = 0.05
Since ;
Pvalue < α
0.025083 < 0.05 ; Reject H0
The confidence interval :
Xbar ± Tcritical * s/sqrt(n)
Tcritical at 95% = 1.761 ;
margin of error = 1.761 * 226.1/sqrt(15) = 102.805
Lower boundary : (5625.1 - 102.805) = 5522.295
(5522.295 ; ∞)
The hypothesized mean does not occur within the constructed confidence boundary. HENCE, there is significant eveidbce to support the claim that the true mean life of a biomedical device is greater than 5500
