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3 years ago

7

Find the volume of the sphere.

1 answer:

Marat540 [252]3 years ago

6 0

Answer:

11/21

Step-by-step explanation:

V = 4/3 $\pi$ $r^{3}$

V = 4/3(22/7) $1/2^{3}$

V = 4/3(22/7)(1/8) = 88/168 or 11/21

It's kind of confusing because of the exponents and fractions but basically the first line if the equation for finding the volume of a sphere. The second is replacing the variables like pie and radius to the given numbers. The last line is all the factions multiplied.

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Use the given x-values to identify solutions as ordered pairs to the equation y = −7x + 4 for x = 1, 3, 5, and 7.

STatiana [176]

Im not sure, maybe determined answer B

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4 years ago

Given circle and circle with radii of 6cm and 4cm respectively.

noname [10]

Solution :

Given that :

The radius of circle A = 6 cm

The radius of circle C = 4 cm

In circle A

$\angle EAF = \theta = 140^\circ$

The length of arc EF = $2 \pi r \times \frac{\theta}{360^\circ}$

$=2 \times 3.14 \times 6 \times \frac{140^\circ}{360^\circ}$

= 14.653 cm

In circle C

$\angle GCH = \theta = 140^\circ$

The length of arc GH = $2 \pi r \times \frac{\theta}{360^\circ}$

$=2 \times 3.14 \times 4 \times \frac{140^\circ}{360^\circ}$

= 9.769 cm

Therefore,

The length of EF is 14.653 cm

The length of GH is 9.769 cm

The length of EF is 1.5 times the length of GH

i.e. 14.653 = 1.5 x 9.769

14.653 = 14.653

Hence proved.

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4 years ago

Compute the simple interest earned on a $5000 deposit earning *.%% for 120 days

tatyana61 [14]

416.6% earning.
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8 0

3 years ago

Read 2 more answers

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uysha [10]

SW= MP

RV=DL

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like number 1: SW is similar MP because then are similar but just differ letters hope that helps.

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3 years ago

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

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

(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