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3 years ago

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You have two rectangular prisms (Prism A and Prism B). Prism A has the following dimensions: length x, width y, and height z. Pr

ism B is made by multiplying each dimension of Prism A by a factor of m, where m >0.
(a) Write a paragraph proof to show that the surface area of Prism B is 2 times the surface area of Prism A.

(b) Write a paragraph proof to show that the volume of Prism B is 3 times the volume of Prism A.

1 answer:

noname [10]3 years ago

3 0

Answer:

Part A) The surface area of prism B is equal to the surface area of prism A multiplied by the scale factor (m) squared

Part B) The Volume of prism B is equal to the Volume of prism A multiplied by the scale factor (m) elevated to the cube

Step-by-step explanation:

Part A) we know that

The scale factor is equal to m

The surface area of the prism is equal to

$S=2B+Ph$

where

B is the area of the base

P is the perimeter of the base

h is the height of the prism

we have

Prism A

$B=xy\ units^{2}$

$P=2(x+y)\ units$

$h=z\ units$

substitute

$SA=[2(xy)+2(x+y)z]\ units^{2}$

Prism B

$B=(mx)(my)=(xy)m^{2}\ units^{2}$

$P=2(mx+my)=2m(x+y)\ units$

$h=mz\ units$

substitute

$SB=[2(xym^{2})+2m(x+y)mz]\ units^{2}$

$SB=[2(xym^{2})+2m^{2}(x+y)z]\ units^{2}$

$SB=m^{2}[2(xy)+2(x+y)z]\ units^{2}$

therefore

The surface area of prism B is equal to the surface area of prism A multiplied by the scale factor (m) squared

Part B) we know that

The volume of the prism is equal to

$V=Bh$

where

B is the area of the base

h is the height of the prism

we have

Prism A

$B=xy\ units^{2}$

$h=z\ units$

substitute

$VA=[(xyz]\ units^{3}$

Prism B

$B=(mx)(my)=(xy)m^{2}\ units^{2}$

$h=mz\ units$

substitute

$VB=[(xym^{2})mz]\ units^{3}$

$VB=[(xyzm^{3})]\ units^{3}$

therefore

The Volume of prism B is equal to the Volume of prism A multiplied by the scale factor (m) elevated to the cube

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Alona [7]

Answer:

Step-by-step explanation:

From the question we are told that

The first sample size is $n_1 = 1000$

The second sample size is $n_2 = 6000$

The number that had significant outside activity in the sample with ALL is $k_1 = 700$

The number that had significant outside activity in the sample without ALL is $k_2 = 5000$

Considering question a

The percentage of children with ALL have significant social activity outside the home when younger is mathematically represented as

$\^ p_1 = \frac{700}{1000} * 100$

=> $\^ p_ 1 = 0.7 = 70\%$

Considering question b

The percentage of children without ALL have significant social activity outside the home when younger is mathematically represented as

$\^ p_2 = \frac{5000}{6000}$

=> $\^ p_ 2 = 0.83$

Generally the sample odds ratio for significant social activity outside the home when younger, comparing the groups with and without ALL is mathematically represented as

$r = \frac{\* p _1}{ \^ p_2 }$

=> $r = \frac{0.7}{ 0.83 }$

=> $r = 0.141$

Considering question c

From the question we are told the confidence level is 95% , hence the level of significance is

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=> $\alpha = 0.05$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

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Generally the lower limit of the 95% confidence interval for the population odds ratio for significant social activity outside the home when younger, comparing the groups with and without ALL is mathematically represented as

$a = e^{ln ( r ) - Z_{\frac{\alpha }{2}} \sqrt{ [ \frac{1}{ k_1 } ] + [ \frac{1}{ c_1 } ] + [\frac{1}{k_2} ] + [\frac{1}{ c_2 } ] } }$

Here $c_1 \ and \ c_2$ are the non-significant values i.e people that did not play outside when they were young in both samples

The values are

$c_1 = 1000 - 700 = 300$

and $c_2 = 6000 - 5000$

=> $c_2 = 1000$

=> $a = e^{ln ( 0.141 ) - 1.96 \sqrt{ [ \frac{1}{ 700 } ] + [ \frac{1}{ 1000} ] + [\frac{1}{5000} ] + [\frac{1}{ 300 } ] } }$

=> $a = 0.1212$

Generally the upper limit of the 95% confidence interval for the population odds ratio for significant social activity outside the home when younger, comparing the groups with and without ALL is mathematically represented as

$b = e^{ln ( 0.141 ) + 1.96 \sqrt{ [ \frac{1}{ 700 } ] + [ \frac{1}{ 1000} ] + [\frac{1}{5000} ] + [\frac{1}{ 300 } ] } }$

$b = 0.1640$

Generally the 95% confidence interval for the population odds ratio for significant social activity outside the home when younger, comparing the groups with and without ALL is

$95\% CI = [ 0.1212 , 0.1640 ]$

Generally looking and the confidence interval obtained we see that it is less that 1 hence this means that there is a greater odd of developing ALL in groups with insignificant social activity compared to groups with significant social activity

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