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3 years ago

54.85 rounded to the nearest 1/10th percent

Mathematics

1 answer:

Iteru [2.4K]3 years ago

8 0

Answer:

54.9

Step-by-step explanation:

You always round up if its 0.5 or anything .5

54.58 broken down:

The 5 is the Tens place

The 4 is the Ones place

The 8 is the Tenth place (what the question is asking for)

The 5 is the Hundredths place.

:) Hope this helps! Please mark brainliest! :)

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Suppose U= {1, 2, 3, 4, 5, 6, 7, 8} is the universal set and P= {1, 2, 3, 4}. What is P?

Digiron [165]

Did you mean P' ? The letter P with a tick mark after it. If so, then start with set U and erase any number you find in set P

We start with
{1,2,3,4,5,6,7,8}
and we erase the following: 1,2,3,4
since those numbers are in set P

P = {1,2,3,4}
P ' = {5, 6, 7, 8}

The set P' is the complement of set P. It is, in a sense, the opposite of set P. It contains everything that P doesn't have.

4 0

3 years ago

60000000000000000000+33333333333333333566566

nasty-shy [4]

Your answer is 3.3393333e+22

8 0

3 years ago

In a packet of 40 skittles, 30% are red. How many skittles are red

kifflom [539]

Answer:

12 skittles are red

Step-by-step explanation:

and easy way to do this on the calculator is 40*0.3 which will give you the number that is the 30%.

6 0

3 years ago

Read 2 more answers

The first term of an Ap is - 504 while the sum of its 9 terms is-126 find Sum of its 30 terms

frozen [14]

Answer:

The sum of its 30 terms is 38167.5

Step-by-step explanation:

Given:

The First term in the AP is -504

The Sum of its 9 terms is -126

To Find:

The sum of its 30 terms = ?

Solution:

The sum of n terms of an AP:

$S_n = (\frac{n}{2} ) [ 2 a_1 + ( n - 1 ) d ]$

The sum of 9 terms of an AP:

$S_9 = (\frac{9 }{2} ) [ 2(-504) + ( 9 - 1 )d ]$

$S_9 = (4.5 )[ 2 (-504)+ ( 8 ) d ]$

$(4.5 )(2)(-504) + (4.5) 8 d = - 126$

(-4536) +36d = -126

36 d = -126+4536

36 d= 4410

$d= \frac{4410}{36}$

d = 122.5

The sum of its 30 terms is

$S_{30} = ( \frac{30 }{2 }) [ 2 (-504) + ( 30-1)(122.5) ]$

$S_{30} =(15) [ 2 (-504) + ( 29)(122.5) ]$

$S_{30} = [ 2 (-504)(15) + ( 29)(122.5)(15) ]$

$S_{30} = [ -15120 + 53287.5 ]$

$s_{30} = 38.167.5$

6 0

4 years ago

The enrollments of all 13 public universities in the state of ohio are listed below. college enrollment university of akron 25,6

bekas [8.4K]

Answers:

(a) The enrollment data is a population
(b) Mean: 22,150.62
(c) Median: 18,670
(d) Range: 57,776
(e) Standard Deviation: 14,258.49

Explanations:

(a) As we noticed in the problem, the enrollment data is used for the study of the number of enrollments in all of the public universities in Ohio. Since the given enrollment data contains the number of enrollments in all of the public univeristies in Ohio, the enrollment data is a population.

(b) To compute the mean, we get the sum of all the data points and divide it by the number of data points.

Since,

Sum of all data points = 1725 + 4344 + 14655 + 15693 + 17338 + 18641 18670 + 20251 + 20839 + 25612 + 34187 + 36502 + 59,501
Sum of all data points = 287,958

and the number of data points = 13,

mean = 287,958 ÷ 13 = 22,150.62

(c) The median is computed when the data points are arranged from least to greatest and it is equal to either:

>> the middle number - if the number of data points is an odd number, or
>> the average of 2 middle numbers - if the number of data points is an even number

In the enrollment data, when the data points are arranged from lowest to highest, the following is the result:

1725, 4344, 14655, 15693, 17338, 18641, 18670, 20251, 20839, 25612, 34187, 36502, 59,501

Since the number of data points is 13 and 13 is an odd number, we find the middle number, which is the 7th lowest number = 18670.

Hence, the median is 18670.

(d) The range of the data is the difference between the highest data point and the lowest data point.

Since the highest data point is 59,501, the lowest data point is 1725 and their difference is 59,501 - 1725 = 57,776. The range of the enrollment data is 57,776.

(e) To compute for standard deviation,

>> First, we compute the squares of all the data points and find their sum:

Sum of the squares of data points = 1725² + 4344² + 14655² + 15693² + 17338² + 18641² + 18670² + 20251² + 20839² + 25612² + 34187² + 36502² + 59,501²

Sum of the squares of data points = 9,021,404,700

>> Then, we compute the average of the squares of all data points by dividing the sum of the squares of all data points by the number of data points:

Average of the squares
= Sum of the squares of data points ÷ number of data points
= 9,021,404,700 ÷ 13
Average of the squares = 693,954,207.69231
 
>> Finally, the standard deviation is the square root of the difference of the average of squares and the square of the average of the data points:

$\sigma = \sqrt{E(X^2) -(E(X))^2}$

where:

$\sigma = \text{standard deviation}
\\ E(X) = \text{average of all data points} = 22,150.62
\\ E(X^2) = \text{average of the squares of all data points} \approx 693,954,207.69231
$

So, the standard deviation is given by

$\sigma = \sqrt{E(X^2) -(E(X))^2}
\\ \sigma = \sqrt{693,954,207.69231 -(22,150.62)^2}
\\ \boxed{\sigma \approx 14,258.49}
$

Therefore, the standard deviation is 14,258.49.

3 0

3 years ago