Which of the following is not a solution of the equation y=3x-4?

Which activity is 6 times faster than the fastest rowing speed?

Debora [2.8K]

The answer is Luge. And I searched up your question and found the whole practice sheet with the answer key. Just put the question on google and you'll find it.

Hope this helps!

6 0

3 years ago

Read 2 more answers

60 % of 80 ASAP pls bruuh

Sati [7]

Answer:

48

Step-by-step explanation:

just ask siri

7 0

4 years ago

The port of South Louisiana, located along 54 miles of the Mississippi River between New Orleans and Baton Rouge, is the largest

Ksenya-84 [330]

Answer:

a) 0.7287

b) 0.9663

c) 0.237

d) 3.65 tons of cargo per week or more that will require the port to extend its operating hours.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 4.5 million tons of cargo per week

Standard Deviation, σ = 0 .82 million tons

We are given that the distribution of number of tons of cargo handled per week is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P( port handles less than 5 million tons of cargo per week)

P(x < 5)

$P( x < 5) = P( z < \displaystyle\frac{5 - 4.5}{0.82}) = P(z < 0.609)$

Calculation the value from standard normal z table, we have,

$P(x < 5) =0.7287= 72.87\%$

b) P( port handles 3 or more million tons of cargo per week)

$P(x \geq 3) = P(z \geq \displaystyle\frac{3-4.5}{0.82}) = P(z \geq −1.82926)\\\\P( z \geq −1.82926) = 1 - P(z < -1.829)$

Calculating the value from the standard normal table we have,

$1 - 0.0337 = 0.9663 = 96.63\%\\P( x \geq 3) = 96.63\%$

c)P( port handles between 3 million and 4 million tons of cargo per week)

$P(3 \leq x \leq 4) = P(\displaystyle\frac{3 - 4.5}{0.82} \leq z \leq \displaystyle\frac{4-4.5}{0.82}) = P(-1.829 \leq z \leq -0.609)\\\\= P(z \leq -0.609) - P(z < -1.829)\\= 0.271-0.034 = 0.237= 23.7\%$

$P(3 \leq x \leq 4) = 23.7\%$

d) P(X=x) = 0.85

We have to find the value of x such that the probability is 0.85.

P(X > x)

$P( X > x) = P( z > \displaystyle\frac{x - 4.5}{0.82})=0.85$

$= 1 -P( z \leq \displaystyle\frac{x - 4.5}{0.82})=0.85$

$=P( z \leq \displaystyle\frac{x - 4.5}{0.82})=0.15$

Calculation the value from standard normal z table, we have,

$P( z \leq -1.036) = 0.15$

$\displaystyle\frac{x - 4.5}{0.82} = -1.036\\x = 3.65$

Thus, 3.65 tons of cargo per week or more that will require the port to extend its operating hours.

8 0

3 years ago

Jamie drives to the dance to pick up his younger brother Mike. Because of the dance's

Vika [28.1K]

Answer:

Because of traffic, she averaged only 15 miles per hour on the way there but ... If the total travel time was 2 hours how long does it take her to drive to ...

Step-by-step explanation:

5 0

3 years ago

The time it takes for a planet to complete its orbit around a particular star is called the? planet's sidereal year. The siderea

BartSMP [9]

Answer:

(a) See below

(b) r = 0.9879

(c) y = -12.629 + 0.0654x

(d) See below

(e) No.

Step-by-step explanation:

(a) Plot the data

I used Excel to plot your data and got the graph in Fig 1 below.

(b) Correlation coefficient

One formula for the correlation coefficient is

$r = \dfrac{\sum{xy} - \sum{x} \sum{y}}{\sqrt{\left [n\sum{x}^{2}-\left (\sum{x}\right )^{2}\right]\left [n\sum{y}^{2} -\left (\sum{y}\right )^{2}\right]}}$

The calculation is not difficult, but it is tedious.

(i) Calculate the intermediate numbers

We can display them in a table.

x y xy x² y²

36 0.22 7.92 1296 0.05

67 0.62 42.21 4489 0.40

93 1.00 93.00 20164 3.46

433 11.8 5699.4 233289 139.24

887 29.3 25989.1 786769 858.49

1785 82.0 146370 3186225 6724

2797 163.0 455911 7823209 26569

3675 248.0 911400 13505625 61504

9965 537.81 1545776.75 25569715 95799.63

(ii) Calculate the correlation coefficient

$r = \dfrac{\sum{xy} - \sum{x} \sum{y}}{\sqrt{\left [n\sum{x}^{2}-\left (\sum{x}\right )^{2}\right]\left [n\sum{y}^{2} -\left (\sum{y}\right )^{2}\right]}}\\\\= \dfrac{9\times 1545776.75 - 9965\times 537.81}{\sqrt{[9\times 25569715 -9965^{2}][9\times 95799.63 - 537.81^{2}]}} \approx \mathbf{0.9879}$

(c) Regression line

The equation for the regression line is

y = a + bx where

$a = \dfrac{\sum y \sum x^{2} - \sum x \sum xy}{n\sum x^{2}- \left (\sum x\right )^{2}}\\\\= \dfrac{537.81\times 25569715 - 9965 \times 1545776.75}{9\times 25569715 - 9965^{2}} \approx \mathbf{-12.629}\\\\b = \dfrac{n \sum xy - \sum x \sum y}{n\sum x^{2}- \left (\sum x\right )^{2}} - \dfrac{9\times 1545776.75 - 9965 \times 537.81}{9\times 25569715 - 9965^{2}} \approx\mathbf{0.0654}\\\\\\\text{The equation for the regression line is $\large \boxed{\mathbf{y = -12.629 + 0.0654x}}$}$

(d) Residuals

Insert the values of x into the regression equation to get the estimated values of y.

Then take the difference between the actual and estimated values to get the residuals.

x y Estimated Residual

36 0.22 -10 10

67 0.62 -8 9

93 1.00 -7 8

142 1.86 -3 5

433 11.8 19 - 7

887 29.3 45 -16

1785 82.0 104 -22

2797 163.0 170 - 7

3675 248.0 228 20

(e) Suitability of regression line

A linear model would have the residuals scattered randomly above and below a horizontal line.

Instead, they appear to lie along a parabola (Fig. 2).

This suggests that linear regression is not a good model for the data.

4 0

3 years ago