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2 years ago

Solve the following system of equations graphically. x = -5 y = -6

Mathematics

1 answer:

zhuklara [117]2 years ago

8 0

The solution to the system of equations is (-5,-6)

<h3>How to solve the system of equations?</h3>

The system of equations is given as:

x = -5

y = -6

Next, we plot both equations on a coordinate plane (see attachment)

From the attached graph, both lines intersect at:

(x,y) = (-5,-6)

Hence, the solution to the system of equations is (-5,-6)

Read more about system of equations at:

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Plsss help quick three question

yKpoI14uk [10]

Answer:

1) A

2) min; min; max; max

3) y = x² + 5x - 3

Step-by-step explanation:

f(x) = x² + 2(x)(5) + 5² - 5² + 24

f(x) = (x + 5)² - 25 + 24

f(x) = (x + 5)² - 1

In ax² + bx + c,

if a > 0, it's a min

if a < 0, it's a max

y = ax² + bx + c

Using (0,-3)

-3 = a(0)² + b(0) + c

c = -3

y = ax² + bx - 3

Using (1,3)

3 = a + b - 3

a + b = 6

Using (-1,-7)

-7 = a(-1)² + b(-1) - 3

-7 + 3 = a - b

a - b = -4

b = a + 4

a + (a + 4) = 6

2a = 2

a = 1

b = 5

y = x² + 5x - 3

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3 years ago

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3 years ago

Define z_alpha to be a z-score with an area of alpha to the right. For Example: z_0.10 means P(Z > z_0.10) = 0.10. We would a

Reptile [31]

Answer:

a) P(-z_0.025 < Z < z_0.025)

For this case we want a quantile that accumulates 0.025 of the area on the tails of the normal standard distribution, and for this case we can calculate the z value with the following excel codes:

"=NORM.INV(0.025,0,1)"

And for this case the two values are : $z_{crit}= \pm 1.96$

b) P(-z_{\alpha/2} < Z < z_{\alpha/2})

For this case we want a quantile that accumulates $\alpha/2$ of the area on the tails of the normal standard distribution, and for this case we can calculate the z value with the following excel codes:

"=NORM.INV(alpha/2,0,1)"

c) For this case we want to find a value of z that satisfy:

P(Z > z_alpha) = 0.05.

And we can use the following excel code:

"=NORM.INV(0.95,0,1)"

And we got $z_{\alpha/2}=1.64$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Part a

P(-z_0.025 < Z < z_0.025)

For this case we want a quantile that accumulates 0.025 of the area on the tails of the normal standard distribution, and for this case we can calculate the z value with the following excel codes:

"=NORM.INV(0.025,0,1)"

And for this case the two values are : $z_{crit}= \pm 1.96$

Part b

P(-z_{\alpha/2} < Z < z_{\alpha/2})

"=NORM.INV(alpha/2,0,1)"

Part c

For this case we want to find a value of z that satisfy:

P(Z > z_alpha) = 0.05.

And we can use the following excel code:

"=NORM.INV(0.95,0,1)"

And we got $z_{\alpha/2}=1.64$

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4 years ago