All categories

3 years ago

Any help please, confused‍♀️

Mathematics

2 answers:

vivado [14]3 years ago

6 0

Use Trigonometry and reread the question and you will understand soon

IceJOKER [234]3 years ago

6 0

Sin 60 = opposite/hypotenuse

opposite is 30. Hypotenuse is what we are looking for. Use algebra:

hypotenuse = opposite/sin60

hypotenuse = 30/(sqrt(3)/2)
hypotenuse = 60/sqrt(3) = 60sqrt(3)/3 = 20sqrt(3)

So the belt is 20sqrt(3) or 34.6 foot

You might be interested in

.) A population of values has a normal distribution with μ=27.5 and σ=71.5. You intend to draw a random sample of size n=180.

Paraphin [41]

Answer: a) 27.5, b) 5.33.

Step-by-step explanation:

Since we have given that

$\mu =27.5\\\\\sigma =71.5$

and n = 180

We need to calculate the sample mean and sample standard deviation.

So, the mean of the distribution of sample means is

$\mu = \mu_x=27.5$

The standard deviation of the distribution of sample means is

$s=\dfrac{\sigma}{\sqrt{n}}=\dfrac{71.5}{\sqrt{180}}=5.329=5.33$

Hence, a) 27.5, b) 5.33.

3 0

3 years ago

Johnstone Rent-A-Car John has a fixed rate of $49 weekly pay is $14 per week for the optional insurance and he spends $12 weekly

amid [387]

Answer:

49+14+12= 75

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

If 2.5 mol of dust particles were laid end to end along the equator, how many times would they encircle the planet? The circumfe

Natalka [10]

Answer:

They encircle the planet $3.76\times 10^{11}$ times.

Step-by-step explanation:

Consider the provided information.

We have 2.5 mole of dust particles and the Avogadro's number is $6.022\times 10^{23}$

Thus, the number of dust particles is:

$2.5\times 6.022\times 10^{23}=15.055\times 10^{23}$

Diameter of a dust particles is 10μm and the circumference of earth is 40,076 km.

Convert the measurement in meters.

Diameter: $10\mu m\times \frac{10^{-6}m}{\mu m} =10^{-5}m$

If we line up the particles the distance they could cover is:

$15.055\times 10^{23}\times 10^{-5}=15.055\times 10^{18}=1.5055\times 10^{19}$

Circumference in meters:

$40,076km\times \frac{1000m}{1km}=40,076,000 m$

Therefore,

$\frac{1.5055\times 10^{19}}{40,076,000} = 3.76\times 10^{11}$

Hence, they encircle the planet $3.76\times 10^{11}$ times.

8 0

3 years ago

Write the product in standard form. ( 9 + 5i)( 9 + 8i)

disa [49]

( 9 + 5i)( 9 + 8i)
= 9(9+8i)+5i(9+8i)
= 81+72i+45i+40i^2
= 81+117i+40(-1)
= 81-40+117i
= 41+117i. Answer

8 0

4 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

3 years ago