Answer:
(x, y) = (1, -1)
Step-by-step explanation:
We'll write these equations in general form, then solve using the cross-multiplication method.
43x +67y +24 = 0
67x +43y -24 = 0
∆1 = (43)(43) -(67)(67) = -2640
∆2 = (67)(-24) -(43)(24) = -2640
∆3 = (24)(67) -(-24)(43) = 2640
These go into the relations ...
1/∆1 = x/∆2 = y/∆3
x = ∆2/∆1 = -2640/-2640 = 1
y = ∆3/∆1 = 2640/-2640 = -1
The solution is (x, y) = (1, -1).
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<em>Additional comment</em>
The cross multiplication method isn't taught everywhere. The attachment explains a bit about it. Our final relationship changes the order of the fractions to 1, x, y from x, y, 1. That way, we can use the equation coefficients in their original general-form order. (The fourth column in the 2×4 array of coefficients is a repeat of the first column.)
Answer:
2.8592933268
Step-by-step explanation:
tan(72) = 8.8/x
=> x = 8.8/tan(72) = 2.859293326849
Hi there!
First, you do first 4x38.15=152.60 then you do 0.12.50x380.50=47.56 then you add total of the result that is how much she own the rental company.
sorry my finger touch the reported button sorry about that.
Hope it was helpful
Answer:
0.9963 is the probabilities that mean number of daily activity of 6 lean people exceed 410 minutes.
Step-by-step explanation:
Solution:
6 Mildly obese people exceed 410 minutes.
Given:
Mean = µ = 376
Standard deviation = ∂ = 70
n = 6
p (x> 410) = p (z > x - µ/ ∂ / √n)
= p (z > 410 – 376 / 70 /√6 )
= p (z > 1.185)
= 1 – p (z < 1.185)
= 1 – 0.38199
= 0.618
0.618 is the probability that the mean number of minutes of daily activity of 6 mildly obese people exceed 410 minutes.
6 lean people exceed 410 minutes:
Given:
Mean = µ = 526
Standard deviation = ∂ = 106
n = 6
=p(x > 410)
= p (z > x - µ/ ∂ /√n)
=p (z > (410 – 526 ) / 106 / √ 6 )
= p (z > - 2.67)
= p (z <2.68)
= 0.9963
0.9963 is the probabilities that mean number of daily activity of 6 lean people exceed 410 minutes.
