Ask question

Ask question

All categories

1 year ago

7

A silent auction was held for a quilt donated for charity. Fifteen bids that were drawn randomly are listed here. Find the value

of the 45th percentile. $85 $240 $120 $60 $315 $190 $145 $105 $285 $260 $95 $215 $235 $175 $155

1 answer:

Alik [6]1 year ago

8 0

45th percentile means that 45% of the data fall below that number while 55% (that is, 100-45) fall above it.

Now, arranging the number in ascending order;
$60, $85, $95, $105, $120, $145, $155, $175, $190, $215, $235, $240, $260, $285, $325

The total number of counts = 15
Then,
Index = 15*45% = 15*0.45 = 6.75

Round up,
Index = 7

The,
7th number = $155

Therefore, 45th percentile value of the data = $155

You might be interested in

Sharon is 54 inches tall. tree in her backyard is five times as tall as she is. the floor of her treehouse is at a height that i

mafiozo [28]

First to solve this problem you are going to figure out how tall the tree is. So if Sharon is 54 inches and the tree 5 times the height she is you would do 54*5= 270. Next you go to find out how tall the floor of her tree house is. It says the floor is twice the height of her so you would do 54*2=108. Then the problem asks the difference between the top of the tree and the floor of the tree house. So 270-108=162. I hope this helped. :) Brainliest answer?

3 0

1 year ago

Read 2 more answers

A school employs 30 teachers.how many will there be if there is a 10% reduction

Georgia [21]

Whats the question bro?

8 0

1 year ago

Factor the expression 28r+42s-35

SIZIF [17.4K]

Answer:7(4r+6s-5)

Step-by-step explanation:

The GCF is 7

3 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

A random survey was taken in Centerville. People in twenty-five homes were asked how many cars were registered to their househol

Ulleksa [173]

In order to find the mean, you first count how many numbers are there. Then, you add all numbers together and divide them by the total of numbers. In this case, you would add (1+2+1+0+3+4+0+1+1+1+2+2+3+2+3+2+1+4+0+0+2+2+1+1+1), which equals to 40. The total of numbers is 25. You divide 40 by 25, and it would get you 1.6. Therefore, your mean is 1.6.

To calculate the median, you list the numbers from least to greatest, and find the middle number. The list for this survey would be 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 4, 4. The middle number of this list is 1, therefore, your median is 1.

The mode is simply the number that appears the most in this list. There are 4 zeroes, 9 ones, 7 twos, 3 threes, and 2 fours. The most in this list would be 1, because there are 9 of them. Your mode is 1.

Mean - 1.6
Median - 1
Mode - 1

Answer: A. 1.6, 1, 1

8 0

2 years ago

Read 2 more answers