Decompose the number 80

A retangle has a perimeter (p) of 58 inches. The length (L) is one more than 3 times the width (W)

worty [1.4K]

The rectangle has a perimeter P of 58 inches.The length l is one more than 3 times the width w.write and solve a system of linear equations to find the length and width of the rectangle?

Answer:

Length(L)=22 inches

Width(W) = 7 inches

Step-by-step explanation:

GIven:-

Perimeter (p)=58 inches,

Length(L)= one more than 3 times the width(W)

Let, W=x ---------------------------------(equation 1

$L=3x+1$ -----------------------(equation 2)

Here x is unknown and to find the Width(W) we have to find the value of x.

Now,

Perimeter of rectangle(p) = 2 times length(L) + 2 times width(W)

$p=2L+2W$

$p=2(3x+1)+2x$ ----------------(from equation 1)

$58=6x+2+2x$ ----------------(given p=58 inches)

$58=8x+2$

$8x=58-2$

$8x=56$

$x=\frac{56}{8}$

$x=7$ ----------------------(equation 3)

Now substituting the value of equation 3 in equation 2.

$L=3x+1$

$L=(3\times 7)+1$

$L=21+1$

$L=22$

$L=22 inches$

as, $W=x$ -----------------------(from equation 1)

$W=7$ inches -------------------(equation 3)

Therefore, Length(L) = 22 inches and Width(W) = 7 inches.

7 0

3 years ago

In your random sample of 25 Chips Ahoy bags, you find a mean of 985 chips per bag, with a standard deviation of 10 chips. Run a

stealth61 [152]

Answer:

The p -value is less than 0.001.

Step-by-step explanation:

Given information:

Sample size = 25 chips

Sample mean = 985

Sample standard deviation = 10

Let as assume that sample is distributed normally.

The formula for test statistic

$t=\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}$

where,

$\overline{x}$ is sample mean

μ is population mean.

s is sample standard deviation.

n is sample size.

The value of test statistic is

$t=\frac{985-1000}{\frac{10}{\sqrt{25}}}$

$t=-7.5$

The p-value is

$p=2\times p(t$

Therefore the p -value is less than 0.001.

6 0

3 years ago

Allowance method entries

Feliz [49]

Using the Allowance Method, the relevant transactions can be completed in the books of Wild Trout Gallery as follows:

1. Allowance for Doubtful Accounts

Accounts Debit Credit

Jan. 1 Beginning balance $53,800

Jan. 19 Accounts Receivable 2,560

Apr. 3 Accounts Receivable $14,670

July 16 Accounts Receivable 19,725

Nov. 23 Accounts Receivable 4,175

Dec. 31 Accounts Receivable 25,110

Dec. 31 Ending balance $56,500

Dec. 31 Bad Debts Expenses $55,470

Totals $116,005 $116,005

Accounts Receivable

Accounts Debit Credit

Jan. 1 Beginning balance $2,290,000

Jan. 19 Allowance for Doubtful 2,560

Jan. 19 Cash $2,560

Apr. 3 Allowance for Doubtful 14,670

July 16 Allowance for Doubtful 19,725

July 16 Cash 6,575

Nov. 23 Allowance for Doubtful 4,175

Nov. 23 Cash 4,175

Dec. 31 Allowance for Doubtful 25,110

Dec. 31 Sales Revenue 8,020,000

Dec. 31 Cash $8,944,420

Dec. 31 Ending balance $1,299,500

Totals $10,316,735 $10,316,735

3. Expected net realizable value of the accounts receivable as of December 31 = $1,243,000 ($1,299,500 - $56,500)

Allowance for Doubtful Accounts ending balance = $40,100 ($8,020,000 x 0.5%)

Allowance for Doubtful Accounts

Accounts Debit Credit

Jan. 1 Beginning balance $53,800

Jan. 19 Accounts Receivable 2,560

Apr. 3 Accounts Receivable $14,670

July 16 Accounts Receivable 19,725

Nov. 23 Accounts Receivable 4,175

Dec. 31 Accounts Receivable 25,110

Dec. 31 Ending balance $40,100

Dec. 31 Bad Debts Expenses $39,070

Totals $99,605 $99,605

4. a. Bad Debt Expense for the year = $39,070

4.b. Balance for Allowance Accounts = $40,100

4.c. Expected net realizable value of the accounts receivable = $1,259,400 ($1,299,500 - $40,100)

Data Analysis:

Jan. 19 Accounts Receivable $2,560 Allowance for Uncollectible Accounts $2,560

Jan. 19 Cash $2,560 Accounts Receivable $2,560

Apr. 3 Allowance for Uncollectible Accounts $14,670 Accounts Receivable $14,670

July 16 Cash $6,575 Allowance for Uncollectible Accounts $19,725 Accounts Receivable $26,300

Nov. 23 Accounts Receivable $4,175 Allowance for Uncollectible Accounts $4,175

Nov. 23 Cash $4,175 Accounts Receivable $4,175

Dec. 31 Allowance for Uncollectible Accounts $25,110 Accounts Receivable $25,110

Accounts Receivable ending balance = $1,299,500

Allowance for Uncollectible Accounts ending balance = $56,500

Learn more: brainly.com/question/22984282

4 0

2 years ago

HELP ME PLEASE !!!!!!!!!!15 POINTS

USPshnik [31]

C is the correct answer 3.785*2.5=9.4625 then multiply by 1000 and you get 9,462.5

3 0

2 years ago

Read 2 more answers

Plssss help in need of answers soon

vfiekz [6]

My guess is that you're doing the Law of Cosines? You have everything you need for that except the angle theta, which is the thing you need to find. It's set up like this: (8)^2 = (10)^2 + (5)^2 -[2(10)(5)cos A] I used A instead of theta. Doing that math, you have: 64 = 100 + 25 -[ 100 cos A]; 64 = 125 - 100 cos A;
-61 = - 100 cos A; -61 / -100 = cos A; .61 = cos A. Now use your inverse function on your calculator to find cos^-1(.61) and that equals 52.4

5 0

2 years ago

Read 2 more answers