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4 years ago

What is the area of the object above?

Mathematics

2 answers:

Mariana [72]4 years ago

8 0

Answer:

Step-by-step explanation:

First, the object is a composite polygon. Its a shape with a bunch of shapes in it. This one has 2 rectangles in it.

One is 12x8, the other is 6x5.

To get the total area, you have to add the area of the smaller shapes.

(12x8)+(6x5)=area

12x8=96

6x5=30

96+30=126

area=126in^2

Ludmilka [50]4 years ago

3 0

C. 126 is the correct answer

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In a regression analysis involving 30 observations, the following estimated regressionequation was obtained.y^ =17.6+3.8x 1 −2.3

USPshnik [31]

Answer:

(a) There is a significant relationship between y and $x_1, x_2, x_3, x_4$

(b) $SSE_{(x_1 ,x_2 ,x_3 ,x_4) }= 45$

(d) $x_1$ and $x_4$ are significant

Step-by-step explanation:

Given

$y = 17.6+3.8x_1 - 2.3x_2 +7.6x_3 +2.7x_4$ --- estimated regression equation

$n = 30$

$p = 4$ --- independent variables i.e. x1 to x4

$SSR = 1760$

$SST = 1805$

$\alpha = 0.05$

Solving (a): Test of significance

We have:

$H_o :$ There is no significant relationship between y and $x_1, x_2, x_3, x_4$

$H_a :$ There is a significant relationship between y and $x_1, x_2, x_3, x_4$

First, we calculate the t-score using:

$t = \frac{SSR}{p} \div \frac{SST - SSR}{n - p - 1}$

$t = \frac{1760}{4} \div \frac{1805- 1760}{30 - 4 - 1}$

$t = 440 \div \frac{45}{25}$

$t = 440 \div 1.8$

$t = 244.44$

Next, we calculate the p value from the t score

Where:

$df = n - p - 1$

$df = 30 -4 - 1=25$

The p value when $t = 244.44$ and $df = 25$ is:

$p =0$

So:

$p < \alpha$ i.e. $0 < 0.05$

Solving (b): $SSE(x_1 ,x_2 ,x_3 ,x_4)$

To calculate SSE, we use:

$SSE = SST - SSR$

Given that:

$SSR = 1760$ ----------- $(x_1 ,x_2 ,x_3 ,x_4)$

$SST = 1805$

So:

$SSE_{(x_1 ,x_2 ,x_3 ,x_4)} = 1805 - 1760$

$SSE_{(x_1 ,x_2 ,x_3 ,x_4) }= 45$

Solving (c): $SSE(x_2 ,x_3)$

To calculate SSE, we use:

$SSE = SST - SSR$

Given that:

$SSR = 1705$ ----------- $(x_2 ,x_3)$

$SST = 1805$

So:

$SSE_{(x_2,x_3)} = 1805 - 1705$

$SSE_{(x_2,x_3)} = 100$

Solving (d): F test of significance

The null and alternate hypothesis are:

We have:

$H_o :$ $x_1$ and $x_4$ are not significant

$H_a :$ $x_1$ and $x_4$ are significant

For this model:

$y =11.1 -3.6x_2+8.1x_3$

$SSE_{(x_2,x_3)} = 100$

$SST = 1805$

$SSR_{(x_2 ,x_3)} = 1705$

$SSE_{(x_1 ,x_2 ,x_3 ,x_4) }= 45$

$p_{(x_2,x_3)} = 2$

$\alpha = 0.05$

Calculate the t-score

$t = \frac{SSE_{(x_2,x_3)}-SSE_{(x_1,x_2,x_3,x_4)}}{p_{(x_2,x_3)}} \div \frac{SSE_{(x_1,x_2,x_3,x_4)}}{n - p - 1}$

$t = \frac{100-45}{2} \div \frac{45}{30 - 4 - 1}$

$t = \frac{55}{2} \div \frac{45}{25}$

$t = 27.5 \div 1.8$

$t = 15.28$

Next, we calculate the p value from the t score

Where:

$df = n - p - 1$

$df = 30 -4 - 1=25$

The p value when $t = 15.28$ and $df = 25$ is:

$p =0$

So:

$p < \alpha$ i.e. $0 < 0.05$

<em>Hence, we reject the null hypothesis</em>

7 0

3 years ago

Simplify radical 16v16

Andreas93 [3]

Answer:

Step-by-step explanation:

4 0

3 years ago

Help pllzzzz !! .....

zimovet [89]

Answer:

image = A = 35×35

preimage = B = 28×28

Scale factor = image/preimage

$\frac{35}{28} \times \frac{35}{28}$

$\frac{5}{4} \times \frac{5}{4}$

= 25/16

6 0

4 years ago

The displacement of a 500 g mass, undergoing simple harmonic motion, is defined by the function :

Valentin [98]

To solve this problem, we have to write out the simple harmonic motion equation and then use the respective formula to solve. The maximum kinetic energy, maximum potential energy and the maximum mechanical energy is equal to 7.56J.

<h3>Simple Harmonic Motion</h3>

The given equation of the simple harmonic motion is

$x = 3.5sin(\frac{\pi}{2t} + \frac{5\pi}{4})\\$

Data;

ω = π/2
k = 1.254N/m

Solving this

$\frac{dx}{dt} = \theta = -3.5 * \frac{\pi}{2} cos (\frac{\pi }{2}t + \frac{5\pi}{4})\\$

Let's calculate the maximum velocity.

$V_m = \frac{3.5\pi}{2}$

This is only possible when cos θ = -1

The maximum kinetic energy is

$K_m = \frac{1}{2}mv_m^2 = \frac{1}{2}* \frac{500}{1000} * (\frac{7\pi}{4} )^2\\K_m = 7.56J$

The maximum kinetic energy is 7.56J

The maximum potential energy can be calculated as

$\omega^2 = \frac{k}{m} \\k = \omega^2 m \\k = \frac{\pi^2}{4} * \frac{500}{1000}\\ k= 1.254 N/m$

Using the value of spring constant, we can find the maximum potential energy

$P.E = \frac{1}{2}kx^2\\P.E = \frac{1}{2} * 1.234 * 3.5^2\\P.E = 7.56J$

The maximum potential energy is 7.56J

The maximum mechanical energy is equal to the sum of maximum potential energy and the maximum kinetic energy.

$ME = P.E + K.E\\ME = 7.56J$

The reason ME is equal to 7.56J is because when x is at maximum, v = 0 and makes kinetic energy equal to zero.

ME = 7.56J

From the calculations above, the maximum kinetic energy, maximum potential energy and the maximum mechanical energy is equal to 7.56J.

Learn more on simple harmonic motion here;

brainly.com/question/15556430

#SPJ1

3 0

2 years ago

The product of two numbers is 3419 and their gcd is 13. Find the lcm

QveST [7]

16 (LCM)=ab3584=ab
so the LCM is 224

8 0

4 years ago