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1 year ago

The volume of a barrel of oil is calculated as 257,166 cm³.

2 answers:

8 0

What is the digit at the hundred thousands place? It is 2. Is the digit to the right of it less than 5? No, 5 is 5. So replace the digits to the right of 2 with 0, and change the 2 to 3.
Answer: 300,000

qaws [65]1 year ago

7 0

Answer:

The nearest hundred thousand of 257,166 cm³ would be 300,000. Some hundred thousand near 257,166 cm³ would be 100,000; 200,000; 300,000; 400,000. And after you see those number, you would see 300,000 is the closet number to 257,166 and the second closest is 200,000.

Hope this help you :3

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Town Hall is located 4.3 miles directly east of the middle school. The fire station is located 1.7 miles directly north of Town

Maksim231197 [3]

Answer:

2.1 miles

Step-by-step explanation:

you can see a drawing of the situation in the attached picture (you need to draw all the places mentioned and the distance between them),

this way we can form a right triangle between the school and the hospital with measures of sides: 1.7 and 1.2 miles

and the green line is the straight line between the school and the hospital, so you need pythagoras to find the value:

$\sqrt{1.7^2+1.2^2}\\=\sqrt{2.89+1.44}\\=\sqrt{4.33} \\ =2.081$

rounding to the nearest tenth:

2.1 miles

4 0

2 years ago

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

Len [333]

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:

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

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

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

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

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

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

3 0

1 year ago

Before the distribution of certain statistical software, every fourth compact disk (CD) is testedfor accuracy. The testing proce

pishuonlain [190]

Answer:

P(T∩E) = 0.017

Step-by-step explanation:

Since every fourth CD is tested. Thus if T is the event that represents 4 disks being tested,

P(T) = 1/4 = 0.25

Let Fi represent event of failure rate. So from the question,

P(F1) = 0.01 ; P(F2) = 0.03 ; P(F3) =0.02 ; P(F4) = 0.01

Also Let F'i represent event of success rate. And we have;

P(F'1) = 1 - 0.01 = 0.99 ; P(F'2) = 1 - 0.03 = 0.97; P(F'3) = 1 - 0.02 = 0.98; P(F'4) = 1 - 0.01 = 0.99

Since all programs run independently, the probability that all programs will run successfully is;

P(All programs to run successfully) =

P(F'1) x P(F'2) x P(F'3) x P(F'4) =

0.99 x 0.97 x 0.98 x 0.97 = 0.932

Now, that all 4 programs failed will be = 1 - 0.932 = 0.068

Let E be denote that the CD fails the test. Thus P(E) = 0.068

Now, since testing and CD's defection are independent events, the probability that one CD was tested and failed will be =P(T∩E) = P(T) x P(E)= 0.25 x 0.068 = 0.017

8 0

2 years ago

Which expression is equivalent to square root 64a^6

zhenek [66]

I believe the answer is A

4 0

1 year ago

Read 2 more answers

Which statement best describes the use of a simulation to predict the probability that two randomly chosen people will both have

laila [671]

<span>Randomly generate an integer from 1 to 7 two times, and the probability is 1/7 ^2

This is the </span><span>statement that best describes the use of a simulation to predict the probability that two randomly chosen people will both have their birthdays on a Monday.

There are 7 days in a week, so there are 7 choices but only 1 Monday. So, 1/7 is the probability that a person's birthday falls on a Monday.

1st person asked will have 1/7 probability.
2nd person asked will also have 1/7 probability

So, (1/7)</span>² is the probability that both persons will have their birthdays on a Monday.

8 0

1 year ago

Read 2 more answers