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4 years ago

How do you find mean meadian and range?

2 answers:

ozzi4 years ago

4 0

Median in the middle number in an ordered sequence of numbers
Ex. 2,4,6,9,10,11,14

The median is 9 because it is in the middle of the ordered series of numbers

Range in simply the largest number minus the smallest number in an ordered sequence of numbers
Ex. 14 - 2 = 12

ipn [44]4 years ago

4 0

To find the Mean add up all the numbers then divide by how many numbers there are. to find the Range take the highest number in the set and the lowest and subtract them and then the Meadian Put all the numbers in numerical order if theres an odd number mark off till you get to the middle if theres an even number of the set the two middle one you will add them both then subtract by 2.

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Simplify: 5/4 times 22/45

Finger [1]

5/4 x 22/45

110/ 180

11/18

6 0

3 years ago

What is the eighth term in the arithmetic sequence defined by the explicit formula

RideAnS [48]

Answer:

Find out the what is the eighth term in the arithmetic sequence defined by the explicit formula.

$a_{n} = 2n + 7$

To prove

As given the explicit formula for the arithmetic sequence .

$a_{n} = 2n + 7$

Put n = 8

$a_{8} = 2\times 8 + 7$

$a_{8} = 16 + 7$

$a_{8} = 23$

Therefore the eighth term in the arithmetic sequence is 23 .

5 0

4 years ago

Read 2 more answers

Sears editing ‘company is a small editorial services company owned and operated by deloris Sears. On January 31, 20Y1 , the end

Valentin [98]

1. Preparation of the adjusting entries for

Sears editing ‘company

Sears editing ‘company Adjusting Entries

a. Dr Insurance expense $1,800

($7,200-$5,400)

Cr Prepaid insurance $1,800

(To record insurance expense)

b. Dr Supplies expense $1,605

($1980 - $375)

Cr Supplies $1,605

(To record supplies expense)

c. Dr Depreciation expense $6,000

Cr Accumulated depreciation-building $6,000

(To record depreciation expense on building)

d. Dr Depreciation expense $3,000

Cr Accumulated depreciation-equipment $3,000

(To record depreciation expense on equipment)

e. Dr Unearned rent $5,400

($6,750-$1,350)

Cr Rent revenue $5,400

(To record rent earned from unearned rent)

f. Dr Salaries and wages expense $2,900

Cr Salaries and wages payable $2,900

(To record salaries and wages expense)

g. Dr Accounts receivable $18,600

Cr Fees earned $18,600

(To record accrued revenue)

2. Preparation of an adjusted trial balance.

Sears editing ‘Company Adjusted Trial balance

Debit side

Cash $7,500

Account Receiveble $57,000

Prepaid Insurance $5,400

Supplies $375

Land $112,500

Building $150,250

Equipment $135,300

Drawing $15,000

Salaries and wages expense $196,270

Utilities expense $42,375

Advertising expense $22,800

Repair expense $17,250

Depreciation expense-Building $6,000

Depreciation expense-Equipment $3,000

Insurance expense $1,800

Supplies expense $1,605

Miscellaneous expense $6,075

Total Debit side $780,500

Credit side

Accumulated Depreciation-Building $93,550

Accumulated Depreciation-Equipment $100,950

Account payable $12,150

Unearned rents $1,350

Salaries and wages payable $2,900

Capital stock $221,000

Fees earned $343,200

Rent revenue $5,400

Total Credit side $780,500

Learn more here:brainly.com/question/21304366

4 0

3 years ago

A rancher wishes to build a fence to enclose a 2250 square yard rectangular field. Along one side the fence is to be made of hea

Bess [88]

Answer:

The least cost of fencing for the rancher is $1200

Step-by-step explanation:

Let x be the width and y the length of the rectangular field.

Let C the total cost of the rectangular field.

The side made of heavy duty material of length of x costs 16 dollars a yard. The three sides not made of heavy duty material cost $4 per yard, their side lengths are x, y, y. Thus

$C=4x+4y+4y+16x\\C=20x+8y$

We know that the total area of rectangular field should be 2250 square yards,

$x\cdot y=2250$

We can say that $y=\frac{2250}{x}$

Substituting into the total cost of the rectangular field, we get

$C=20x+8(\frac{2250}{x})\\\\C=20x+\frac{18000}{x}$

We have to figure out where the function is increasing and decreasing. Differentiating,

$\frac{d}{dx}C=\frac{d}{dx}\left(20x+\frac{18000}{x}\right)\\\\C'=20-\frac{18000}{x^2}$

Next, we find the critical points of the derivative

$20-\frac{18000}{x^2}=0\\\\20x^2-\frac{18000}{x^2}x^2=0\cdot \:x^2\\\\20x^2-18000=0\\\\20x^2-18000+18000=0+18000\\\\20x^2=18000\\\\\frac{20x^2}{20}=\frac{18000}{20}\\\\x^2=900\\\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\x=\sqrt{900},\:x=-\sqrt{900}\\\\x=30,\:x=-30$

Because the length is always positive the only point we take is $x=30$ . We thus test the intervals $(0, 30)$ and $(30, \infty)$

$C'(20)=20-\frac{18000}{20^2} = -25 < 0\\\\C'(40)= 20-\frac{18000}{20^2} = 8.75 >0$

we see that total cost function is decreasing on $(0, 30)$ and increasing on $(30, \infty)$ . Therefore, the minimum is attained at $x=30$ , so the minimal cost is

$C(30)=20(30)+\frac{18000}{30}\\C(30)=1200$