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deff fn [24]
2 years ago
8

the solution to the system of equation below is (−2, −1). 2x − 3y = −1 11x − 9y = −13 when the first equation is multiplied by −

3, the sum of the two equations is equivalent to 5x = −10.
Mathematics
1 answer:
Evgen [1.6K]2 years ago
6 0
When the first equation is multiplied by -3, the sum of the two equation and any of the original equations will have the same solution as the original equations. Hence 5x = -10 and 11x - 9x = -13 will have a solution of (-2, -1).
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The expected value of the amount of average snowfall for over 30 years is 86.7 inches with a standard deviation of 40.4 inches. To verify if this particular trend continues, we must check the significance value of the amount snowfall for the past four years. Given that the snowfall for past years are as follows: 115.7 inches, 62.9 inches, 168.5 inches, and 135.7 inches. Thus the mean of the sample would be: (115.7 + 62.9 + 168.5 + 135.7)/4 = 120.7 inches. To compute for the z-score, we have z-score = (x – μ) / (σ / √n) where x is the computed/measured value, μ is the expected mean, σ is the standard deviation, and n is the number of samples. Using the information we have, z-score (z) = (120.7 - 86.7) / (40.4/ √4) = 1.68 In order to reject the null hyptohesis our probability value must be less than the significance level of 5%. For our case, since z = 1.68, P-value = 0.093 > 0.05. Therefore, the answer is B.
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Wynn needs to find the center of the data set shown on the dot plot. The dot plot has dots. The dot plot has dots to the left of
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dots.

The dot plot has   11

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center and   11

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Step-by-step explanation:

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2 years ago
In equilateral ∆ABC with side a, the perpendicular to side AB at point B intersects extension of median AM in point P. What is t
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Answer:

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Step-by-step explanation:

Given: ΔABC is equilateral and AB = a

The diagram is given below :

AM is a median , PB ⊥ AB , PM = b

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We get, ∠AMB = 90°. So, by linear pair ∠AMB + ∠PMB = 180° ⇒ ∠PMB = 90°. Also, ∠ABC = 60° and ∠ABP = 90° (given) So, ∠PBM = 30°

Since, AM is perpendicular bisector of BC. So,

MB = \frac{a}{2}

Now in ΔAMB , By using Pythagoras theorem

AB^{2}=AM^{2}+MB^{2}\\AM^{2}=AB^{2}-MB^{2}\\AM^{2}=a^{2}-(\frac{a}{2})^{2}\\AM=\frac{\sqrt{3}\cdot a}{2}

Now, in ΔBMP :

sin\thinspace 30^{o}=\frac{\text{Perpendicular}}{\text{Hypotenuse}}\\\\sin\thinspace 30^{o}=\frac{\text{MB}}{\text{PB}}\\\\PB=\frac{\text{MB}}{\text{sin 30}}\\\\PB=\frac{\frac{a}{2}}{\frac{1}{2}}\implies PB = a\\\\tan\thinspace 30^{o}=\frac{\text{Perpendicular}}{\text{Base}}\\\\tan\thinspace 30^{o}=\frac{\text{MB}}{\text{PM}}\\\\PM=\frac{\text{MB}}{\text{tan 30}}\\\\PM=\frac{\frac{a}{2}}{\frac{1}{\sqrt3}}\implies PM=b= \frac{\sqrt{3}\cdot a}{2}

Perimeter of ABM = AB + PB + PM + AM

\text{Perimeter = }a+a+b+ \frac{\sqrt{3}\cdot a}{2}\\\\=2\cdot a + \frac{\sqrt{3}\cdot a}{2} +\frac{\sqrt{3}\cdot a}{2}\\\\=2\cdot a +\sqrt{3}\cdot a\\\\=(2+\sqrt3})\cdot a

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