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2 years ago

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A standard city block in Manhattan is a rectangle measuring 80 m by 270 m. A resident wants to get from one corner of a block to

the opposite corner of a block that contains a park. She wonders about the difference between cutting across the diagonal through the park compared to going around the park, along the streets.
How much shorter would her walk be going through the park? Round your answer to the nearest meter.

1 answer:

kondor19780726 [428]2 years ago

5 0

Answer:

68.4 meters shorter

Step-by-step explanation:

ok so you want to compare the distance 80 m + 270 m with the diagonal length.

the diagonal length = root ( 80^2 + 270^2 )

diagonal length = root(6400 + 72900)

diagonal = root(79300) =281.602 m

so.

350 - 281.602 = 68.397 meters about 68.4 meters shorter than walking around

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The selection of r objects out of n is done in

$C(n, r)= \frac{n!}{r!(n-r)!}$ many ways.

The total number of selections 10 that we can make from 6+7=13 students is

$C(13,10)= \frac{13!}{3!(10)!}= \frac{13*12*11*10!}{3*2*1*10!}= \frac{13*12*11}{3*2}= 286$
thus, the sample space of the experiment is 286

A.
<span>"The probability that a randomly chosen team includes all 6 girls in the class."

total number of group of 10 which include all girls is C(7, 4), because the girls are fixed, and the remaining 4 is to be completed from the 7 boys, which can be done in C(7, 4) many ways.

</span> $C(7, 4)= \frac{7!}{4!3!}= \frac{7*6*5*4!}{4!*3*2*1}= \frac{7*6*5}{3*2}=35$
<span>
P(all 6 girls chosen)=35/286=0.12

B.
"</span>The probability that a randomly chosen team has 3 girls and 7 boys.<span>"

with the same logic as in A, the number of groups were all 7 boys are in, is

</span> $C(6, 3)= \frac{6!}{3!3!}= \frac{6*5*4*3!}{3!3!}= \frac{6*5*4}{3*2*1}=20$
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C.
"</span>The probability that a randomly chosen team has either 4 or 6 boys.<span>"

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there C(7, 6)*C(6, 4) ,many ways of doing this, because any selection of the boys which can be done in C(7, 6) ways, can be combined with any selection of the girls.

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The answer would be 1,220,000mm.

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There are 1, 000,000 mm in one km.

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A store sells 8 colors of balloons with at least 28 of each color. How many different combinations of 28 balloons can be chosen?

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Answer:

(a) $Selection = 6724520$

(b) $At\ most\ 12 = 6553976$

(c) $At\ most\ 8 = 6066720$

(d) $At\ most\ 12\ red\ and\ at\ most\ 8\ blue = 5896638$

Step-by-step explanation:

Given

$Colors = 8$

$Balloons = 28$ --- at least

Solving (a): 28 combinations

From the question, we understand that; a combination of 28 is to be selected. Because the order is not important, we make use of combination.

Also, because repetition is allowed; different balloons of the same kind can be selected over and over again.

So:

$n => 28 + 8-1$ $= 35$

$r = 28$

$Selection = ^{35}^C_{28$

$Selection = \frac{35!}{(35 - 28)!28!}$

$Selection = \frac{35!}{7!28!}$

$Selection = \frac{35*34*33*32*31*30*29*28!}{7!28!}$

$Selection = \frac{35*34*33*32*31*30*29}{7!}$

$Selection = \frac{35*34*33*32*31*30*29}{7*6*5*4*3*2*1}$

$Selection = \frac{33891580800}{5040}$

$Selection = 6724520$

Solving (b): At most 12 red balloons

First, we calculate the ways of selecting at least 13 balloons

Out of the 28 balloons, there are 15 balloons remaining (i.e. 28 - 13)

So:

$n => 15 + 8 -1 = 22$

$r = 15$

Selection of at least 13 =

$At\ least\ 13 = ^{22}C_{15}$

$At\ least\ 13 = \frac{22!}{(22-15)!15!}$

$At\ least\ 13 = \frac{22!}{7!15!}$

$At\ least\ 13 = 170544$

Ways of selecting at most 12 =

$At\ most\ 12 = Total - At\ least\ 13$ --- Complement rule

$At\ most\ 12 = 6724520- 170544$

$At\ most\ 12 = 6553976$

Solving (c): At most 8 blue balloons

First, we calculate the ways of selecting at least 9 balloons

Out of the 28 balloons, there are 19 balloons remaining (i.e. 28 - 9)

So:

$n => 19+ 8 -1 = 26$

$r = 19$

Selection of at least 9 =

$At\ least\ 9 = ^{26}C_{19}$

$At\ least\ 9 = \frac{26!}{(26-19)!19!}$

$At\ least\ 9 = \frac{26!}{7!19!}$

$At\ least\ 9 = 657800$

Ways of selecting at most 8 =

$At\ most\ 8 = Total - At\ least\ 9$ --- Complement rule

$At\ most\ 8 = 6724520- 657800$

$At\ most\ 8 = 6066720$

Solving (d): 12 red and 8 blue balloons

First, we calculate the ways for selecting 13 red balloons and 9 blue balloons

Out of the 28 balloons, there are 6 balloons remaining (i.e. 28 - 13 - 9)

So:

$n =6+6-1 = 11$

$r = 6$

Selection =

$^{11}C_6 = \frac{11!}{(11-6)!6!}$

$^{11}C_6 = \frac{11!}{5!6!}$

$^{11}C_6 = 462$

Using inclusion/exclusion rule of two sets:

$Selection = At\ most\ 12 + At\ most\ 8 - (12\ red\ and\ 8\ blue)$

$Only\ 12\ red\ and\ only\ 8\ blue\ = 170544+ 657800- 462$

$Only\ 12\ red\ and\ only\ 8\ blue\ = 827882$

$At\ most\ 12\ red\ and\ at\ most\ 8\ blue = Total - Only\ 12\ red\ and\ only\ 8\ blue$

$At\ most\ 12\ red\ and\ at\ most\ 8\ blue = 6724520 - 827882$

$At\ most\ 12\ red\ and\ at\ most\ 8\ blue = 5896638$

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2 years ago