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icang [17]
3 years ago
15

Using the regression formula with a slope = 2,000 and an intercept = 15,000, what would the predicted income be for someone who

has 16 years of education? a $40,000 b $42,000 c $45,000 d $47,000
Mathematics
1 answer:
rodikova [14]3 years ago
8 0

Answer:

d = 47000

Step-by-step explanation:

The exercise is already giving the data we need to calculate this.

First, we have to use the linear regression formula which is:

y = a + bx

where:

a = intercept

b = slope

In this case we already have the slope and intercept, the value of x, will be the number of years reflected here, therefore, the predicted income would be:

y = 2000(16) + 15000

y = 32000 + 15000

y = 47,000

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Tell whether the angles are adjacent or vertical. Then find the value of x.
Gwar [14]

Answer:

x = 128

Step-by-step explanation:

Adjacent angles - Angles right next to each other. They share a common vertex

Vertical angles - angles that are vertically aligned with each other.

The angle that has a measure of 128 degrees and x are vertical angles

If you didn't know vertical angles are congruent so because one angle equals 128 degrees the other equals 128 degrees

x = 128

6 0
3 years ago
Read 2 more answers
Evaluate<br> -3.2 + ] (16.4) – 3 =
Maksim231197 [3]

Answer:

10.2

Step-by-step explanation:

-3.2+(16.4)-3=

13.2-3

10.2

3 0
3 years ago
A data set lists earthquake depths. The summary statistics are
irinina [24]

Answer:

Step-by-step explanation:

The summary of the given statistics data include:

sample size n = 400

sample mean \overline x = 6.86

standard deviation = 4.37

Level of significance ∝ = 0.01

Population Mean \mu = 6.00

Assume that a simple random sample has been selected. Identify the null and alternative​ hypotheses, test​ statistic, P-value, and state the final conclusion that addresses the original claim.

To start with the hypothesis;

The null and the alternative hypothesis can be computed as :

H_o: \mu = 6.00 \\ \\  H_1 : \mu \neq 6.00

The test statistics for this two tailed test can be computed as:

z= \dfrac{\overline x - \mu}{\dfrac{\sigma}{\sqrt {n}}}

z= \dfrac{6.86 - 6.00}{\dfrac{4.37}{\sqrt {400}}}

z= \dfrac{0.86}{\dfrac{4.37}{20}}

z = 3.936

degree of freedom = n - 1

degree of freedom = 400 - 1

degree of freedom = 399

At the level of significance ∝ = 0.01

P -value = 2 × (z < 3.936)  since it is a two tailed test

P -value = 2 × ( 1  - P(z ≤ 3.936)

P -value = 2 × ( 1  -0.9999)

P -value = 2 × ( 0.0001)

P -value =  0.0002

Since the P-value is less than level of significance , we reject H_o at level of significance 0.01

Conclusion: There is sufficient evidence to conclude that the original claim that the mean of the population of earthquake depths is  5.00 km.

7 0
3 years ago
Find the equation for the<br> following parabola.<br> Vertex (2, -1)<br> Focus (2, 3)
Naily [24]

Answer:

(x−2)^2=16(y+1)

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Help please!What is the x-intercept of the line?X=33 Y=-22X=52 Y=-33X=71 Y=-44
tekilochka [14]

x intercept is -5

Explanation

Step 1

find the slope of the line:

when you know 2 points of the line, you can find the slope, by using:

\begin{gathered} \text{slope}=\frac{change\text{ in y}}{\text{change in x}}=\frac{y_2-y_1}{x_2-x_1} \\ \text{where} \\ P1(x_1,y_1) \\ P2(x_2,y_2) \end{gathered}

then,Let

P1(33,-22)

P2(52,-33)

replace,

\begin{gathered} \text{slope}=\frac{change\text{ in y}}{\text{change in x}}=\frac{y_2-y_1}{x_2-x_1} \\ \text{slope}=\frac{-33-(-22)}{52-33}=\frac{-33+22}{19}=\frac{-11}{19} \\ \text{slope}=-\frac{11}{19} \end{gathered}

Step 2

find the equation of the line

y-y_1=slope(x-x_1)\rightarrow equation

let

\begin{gathered} \text{slope}=-\frac{11}{19} \\ P1(33,-22) \end{gathered}

replace,

\begin{gathered} y-y_1=slope(x-x_1)\rightarrow equation \\ y-(-22)=-\frac{11}{19}(x-33) \\ y+22=-\frac{11}{19}x+\frac{363}{19} \\ to\text{ isolate y, subtract 22 in both sides} \\ y+22-22=-\frac{11}{19}x+\frac{363}{19}-22 \\ y=-\frac{11}{19}x-\frac{55}{19}\rightarrow equation\text{ of the line} \end{gathered}

now, we have the equation of the line, to get the x intercetp ( it is when y=0)

replace

\begin{gathered} y=-\frac{11}{19}x-\frac{55}{19}\rightarrow equation\text{ of the line} \\ 0=-\frac{11}{19}x-\frac{55}{19} \\ \text{isolate x} \\ \frac{11}{19}x=-\frac{55}{19} \\ 11x=-\frac{55\cdot19}{19} \\ 11x=-55 \\ \text{divide both sides by 11} \\ \frac{11x}{11}=\frac{-55}{11} \\ x=-5 \end{gathered}

so, the x intercetp is -5.

I hope this helps you

6 0
1 year ago
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