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Kitty [74]
4 years ago
14

Two lines, A and B, are represented by the equations given below!

Mathematics
1 answer:
lara [203]4 years ago
8 0

Answer:

The solutions to the system of equations are:

y=-8,\:x=-4

Thus, option C is true because the point satisfies BOTH equations.

Step-by-step explanation:

Given the system of the equations

\begin{bmatrix}y=x-4\\ y=3x+4\end{bmatrix}

Arrange equation variables for elimination

\begin{bmatrix}y-x=-4\\ y-3x=4\end{bmatrix}

y-3x=4

-

\underline{y-x=-4}

-2x=8

\begin{bmatrix}y-x=-4\\ -2x=8\end{bmatrix}

solve for x

-2x=8

Divide both sides by -2

\frac{-2x}{-2}=\frac{8}{-2}

x=-4

\mathrm{For\:}y-x=-4\mathrm{\:plug\:in\:}x=-4

y-\left(-4\right)=-4

y+4=-4

y=-8

The solutions to the system of equations are:

y=-8,\:x=-4

Thus, option C is true because the point satisfies BOTH equations.

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So the slope is 5 and you know the points, (0,4)
y-4=5(x-0)
y-4 = 5x
y = 5x + 4 is the slope intercept form
standard form is: -5x + y = 4 OR y - 5x = 4
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3 years ago
)) Matt's bill for breakfast at a restaurant was $91. He left an 18% tip. What was the total
gladu [14]

Answer:

$107.38

Step-by-step explanation:

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$91 (th original price) + 15.38 (the tip) = $108.38 (final price)

4 0
3 years ago
Read 2 more answers
He goodness of fit is measured by χ2. This statistic measures the amounts by which the observed values differ from their respect
Murrr4er [49]

Answer:

Step-by-step explanation:

Hello!

Table is attached.

1) Purple stem/short petals.

(o-e)= 220-225= -5

(o-e)²= (-5)²= 25

(o-e)²/e= 25/225= 1/9

2) Green stem/short petals

(o-e)= 210-225= -15

(o-e)²= (-15)²= 225

(o-e)²/e= 225/225= 1

3) Purple stem/ long petals

(o-e)= 231-225= 6

(o-e)²= (6)²= 36

(o-e)²/e= 36/225= 4/25

4) Green stem/ long petals

(o-e)= 239-225= 14

(o-e)²= (14)²= 196

(o-e)²/e= 196/225 (≅ 0.871)

To reach the statistic value you have to add the four values obtained in the last row:

X²= 1/9+1+4/25+196/225= 482/225≅ 2.142

I hope you have a SUPER day!

4 0
4 years ago
"Suppose we are testing the null hypothesis H0: = 20 and the alternative Ha: 20, for a Normal population with = 5. A random samp
mezya [45]

Complete question is;

Suppose we are testing the null hypothesis H0: μ = 20 and the alternative Ha: μ ≠ 20, for a normal population with σ = 5. A random sample of 25 observations are drawn from the population, and we find the sample mean of these observations is x¯ = 17.6. The P-value is closest to:

a. 0.0668.

b. 0.0082.

c. 0.0164.

d. 0.1336

Answer:

Option D

Step-by-step explanation:

We are given;

Null hypothesis; H0: μ = 20

Alternative hypothesis; Ha: μ ≠ 20

Population Standard deviation; σ = 5

Sample size; n = 25

Sample mean; x¯ = 17.6

Let's find the z-score from the formula;

z = (x¯ - μ)/(σ/√n)

z = (17.6 - 20)/(5/√5)

z = - 1.073

From online p-value from z-score calculator attached, using z = -1.073; two tailed hypothesis; significance value of 0.05,we have;

The P-Value is 0.283271.

Looking at the given options, the closest to the p-value is option D

8 0
3 years ago
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