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2 years ago

Optimization-

Mathematics

1 answer:

pishuonlain [190]2 years ago

5 0

Do you want a proof or just the answer?
The answer is: the square (2.5x2.5).

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When the sun has risen 32 degrees above the horizon, Sandy casts a shadow that is 9 feet 2 inches long. How tall is Sandy, to th

Makovka662 [10]

Answer:

The height of Sandy is 68.74 inches.

Step-by-step explanation:

It is given that the sun has risen 32 degrees above the horizon. It means the angle of elevation is 32°.

Let the height of Sandy be x.

Sandy casts a shadow that is 9 feet 2 inches long.

We know that 1 feet = 12 inches

Using this conversion, we get

Shadow = 9 feet 2 inches = 9(12)+2 = 110 inches

In a right angled triangle,

$\tan\theta =\frac{perpendicular}{base}$

In triangle ABC,

$\tan(\angle C)=\frac{AB}{BC}$

$\tan(32^{\circ})=\frac{x}{110}$

Multiply both sides by 110.

$110\tan(32^{\circ})=x$

$68.73562871=x$

$x\approx 68.74$

Therefore the height of Sandy is 68.74 inches.

8 0

2 years ago

Expand (3x+2)^4 and show your work.

AysviL [449]

You have to distribute the exponent of 4 to the whole equation.
so step one is to distribute to the 3
3^4= 81
then we have to distribute to the x
x^4
then to the 2
2^4= 16
81 * x^4 * 16
* = multiplication sign

7 0

2 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

(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

2 years ago

Given below are the graphs of two lines, y=-0.5 + 5 and y=-1.25x + 8 and several regions and points are shown. Note that C is th

zalisa [80]

We have the following equations:

$(1) \ y=-0.5x+5 \\ (2) \ y=-1.25x+8$

So we are asked to write a system of equations or inequalities for each region and each point.

Part a)

Region Example A

$y \leq -0.5x+5 \\ y \leq -1.25x+8$

Region B.

Let's take a point that is in this region, that is:

$P(0,6)$

So let's find out the signs of each inequality by substituting this point in them:

$y \ (?)-0.5x+5 \\ 6 \ (?) -0.5(0)+5 \\ 6 \ (?) \ 5 \\ 6\ \textgreater \ 5 \\ \\ y \ (?) \ -1.25x+8 \\ 6 \ (?) -1.25(0)+8 \\ 6 \ (?) \ 8 \\ 6\ \textless \ 8$

So the inequalities are:

$(1) \ y \geq -0.5x+5 \\ (2) \ y \leq -1.25x+8$

Region C.

A point in this region is:

$P(0,10)$

So let's find out the signs of each inequality by substituting this point in them:

$y \ (?)-0.5x+5 \\ 10 \ (?) -0.5(0)+5 \\ 10 \ (?) \ 5 \\ 10\ \textgreater \ 5 \\ \\ y \ (?) \ -1.25x+8 \\ 10 \ (?) -1.25(0)+8 \\ 10 \ (?) \ 8 \\ 10 \ \ \textgreater \ \ 8$

So the inequalities are:

$(1) \ y \geq -0.5x+5 \\ (2) \ y \geq -1.25x+8$

Region D.

A point in this region is:

$P(8,0)$

So let's find out the signs of each inequality by substituting this point in them:

$y \ (?)-0.5x+5 \\ 0 \ (?) -0.5(8)+5 \\ 0 \ (?) \ 1 \\ 0 \ \ \textless \ \ 1 \\ \\ y \ (?) \ -1.25x+8 \\ 0 \ (?) -1.25(8)+8 \\ 0 \ (?) \ -2 \\ 0 \ \ \textgreater \ \ -2$

So the inequalities are:

$(1) \ y \leq -0.5x+5 \\ (2) \ y \geq -1.25x+8$

Point P:

This point is the intersection of the two lines. So let's solve the system of equations:

$(1) \ y=-0.5x+5 \\ (2) \ y=-1.25x+8 \\ \\ Subtracting \ these \ equations: \\ 0=0.75x-3 \\ \\ Solving \ for \ x: \\ x=4 \\ \\ Solving \ for \ y: \\ y=-0.5(4)+5=3$

Accordingly, the point is:

$\boxed{p(4,3)}$

Point q:

This point is the $x-intercept$ of the line:

$y=-0.5x+5$

So let:

$y=0$

Then

$x=\frac{5}{0.5}=10$

Therefore, the point is:

$\boxed{q(10,0)}$

Part b)

The coordinate of a point within a region must satisfy the corresponding system of inequalities. For each region we have taken a point to build up our inequalities. Now we will take other points and prove that these are the correct regions.

Region Example A

The origin is part of this region, therefore let's take the point:

$O(0,0)$

Substituting in the inequalities:

$y \leq -0.5x+5 \\ 0 \leq -0.5(0)+5 \\ \boxed{0 \leq 5} \\ \\ y \leq -1.25x+8 \\ 0 \leq -1.25(0)+8 \\ \boxed{0 \leq 8}$

It is true.

Region B.

Let's take a point that is in this region, that is:

$P(0,7)$

Substituting in the inequalities:

$y \geq -0.5x+5 \\ 7 \geq -0.5(0)+5 \\ \boxed{7 \geq \ 5} \\ \\ y \leq \ -1.25x+8 \\ 7 \ \leq -1.25(0)+8 \\ \boxed{7 \leq \ 8}$

It is true

Region C.

Let's take a point that is in this region, that is:

$P(0,11)$

Substituting in the inequalities:

$y \geq -0.5x+5 \\ 11 \geq -0.5(0)+5 \\ \boxed{11 \geq \ 5} \\ \\ y \geq \ -1.25x+8 \\ 11 \ \geq -1.25(0)+8 \\ \boxed{11 \geq \ 8}$

It is true

Region D.

Let's take a point that is in this region, that is:

$P(9,0)$

Substituting in the inequalities:

$y \leq -0.5x+5 \\ 0 \leq -0.5(9)+5 \\ \boxed{0 \leq \ 0.5} \\ \\ y \geq \ -1.25x+8 \\ 0 \geq -1.25(9)+8 \\ \boxed{0 \geq \ -3.25}$

It is true

7 0

2 years ago

Which equation can be used to find the nth term in the following sequence?

Annette [7]

Answer:

d bio d if husks on zips by hdjxbxbdibdksoshfbrinrl oh s Yb exhort gnu in dudeneijdidnidbduow9sh

7 0

2 years ago