410 is what percent of 65?, №17886133, 02.06.2020 01:50

Mathematics

Step-by-step answer

P Answered by PhD

15.85%

Step-by-step explanation:

you know you have 65 and you have 410 so you just multiply and try and figure out the ratio and how much percentage is out of it.

Chemistry

The practical limit to ages that can be determined by radiocarbon dating is about 41000-yr-old sample, what percentage of the original 614c atoms remains? (for percentage you can either enter the percentage with a percent sign 65.1% or the percentage as a fraction 0.651 .)

Step-by-step answer

P Answered by Master

1

In percentage, the sample of C-4 remains = 0.7015 %

Explanation:

The Half life Carbon 14 = 5730 year

$t_{1/2}=\frac {ln\ 2}{k}$

Where, k is rate constant

So,

$k=\frac {ln\ 2}{t_{1/2}}$

$k=\frac {ln\ 2}{5730}\ hour^{-1}$

The rate constant, k = 0.000120968 year⁻¹

Time = 41000 years

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

[A_t] is the concentration at time t

[A_0] is the initial concentration

So,

$\frac {[A_t]}{[A_0]}=e^{-0.000120968\times 41000}$

$\frac {[A_t]}{[A_0]}=0.007015$

In percentage, the sample of C-4 remains = 0.7015 %

Chemistry

The practical limit to ages that can be determined by radiocarbon dating is about 41000-yr-old sample, what percentage of the original 614c atoms remains? (for percentage you can either enter the percentage with a percent sign 65.1% or the percentage as a fraction 0.651 .)

Step-by-step answer

P Answered by Specialist

1

In percentage, the sample of C-4 remains = 0.7015 %

Explanation:

The Half life Carbon 14 = 5730 year

$t_{1/2}=\frac {ln\ 2}{k}$

Where, k is rate constant

So,

$k=\frac {ln\ 2}{t_{1/2}}$

$k=\frac {ln\ 2}{5730}\ hour^{-1}$

The rate constant, k = 0.000120968 year⁻¹

Time = 41000 years

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

[A_t] is the concentration at time t

[A_0] is the initial concentration

So,

$\frac {[A_t]}{[A_0]}=e^{-0.000120968\times 41000}$

$\frac {[A_t]}{[A_0]}=0.007015$

In percentage, the sample of C-4 remains = 0.7015 %

Business

The Hill Company reported the following results: Year 3 Year 2 Year 1 Income Statement Revenue 10,972 11,598 10,470 Cost of Goods Sold 8,942 8,767 7,901 Selling, General & Admin. Expenses 2,470 2,611 2,479 Interest expense 76 80 28 Net Income (516) 140 62 Balance Sheet Assets Cash 1,354 1,316 1,880 Prepaid expenses 202 522 125 Accounts receivable 375 250 231 Inventory 745 698 455 Property & equipment (net) 20,464 18,810 17,727 Total Assets 23,140 21,596 20,418 Liabilities Accounts payable 2,824 743 678 Unredeemed gift cards 410 850 636

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P Answered by PhD

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Business

The Hill Company reported the following results: Year 3 Year 2 Year 1 Income Statement Revenue 10,972 11,598 10,470 Cost of Goods Sold 8,942 8,767 7,901 Selling, General & Admin. Expenses 2,470 2,611 2,479 Interest expense 76 80 28 Net Income (516) 140 62 Balance Sheet Assets Cash 1,354 1,316 1,880 Prepaid expenses 202 522 125 Accounts receivable 375 250 231 Inventory 745 698 455 Property & equipment (net) 20,464 18,810 17,727 Total Assets 23,140 21,596 20,418 Liabilities Accounts payable 2,824 743 678 Unredeemed gift cards 410 850 636

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Biology

Almost 50 percent of water withdrawn in the us, is used up by thermoelectric power plants. the table below shows the amount of water withdrawn in the u.s., in billion gallons per year, between 1950 and 2005. year amount of water withdrawn in the us (in billion gallons per year) 1950 180 1965 340 1975 420 1990 410 2005 410 based on the table, which of these conclusions is most likely correct? the number of thermal power plants decreased between 1950 and 1975. the number of thermal power plants increased between 1950 and 1975. thermoelectric power

Step-by-step answer

P Answered by Master

18

The number of thermal power plants increased between 1950 and 1975.

Explanation:

Here, the table that shows the amount of water withdrawn in billion gallons per year, between 1950 and 2005,

Year 1950 1965 1975 1990 2005

Amount 180 340 420 410 410

Since, 420 > 180,

Also, 50 percent of water withdrawn is used up by thermometric power plants.

Thus, thermal power plants in 1975 > thermal power plant in 1950

Therefore, the number of thermal power plants increased between 1950 and 1975.

Second option is correct.

Note : first option is incorrect.

Since, by the above table,

The least amount of water used by power plant is 180 billion gallons per year,

Thus, third option is incorrect,

Now, the table does not represents energy produced,

Fourth option can not be said correct.

Biology

Almost 50 percent of water withdrawn in the us, is used up by thermoelectric power plants. the table below shows the amount of water withdrawn in the u.s., in billion gallons per year, between 1950 and 2005. year amount of water withdrawn in the us (in billion gallons per year) 1950 180 1965 340 1975 420 1990 410 2005 410 based on the table, which of these conclusions is most likely correct? the number of thermal power plants decreased between 1950 and 1975. the number of thermal power plants increased between 1950 and 1975. thermoelectric power

Step-by-step answer

P Answered by Master

18

The number of thermal power plants increased between 1950 and 1975.

Explanation:

Here, the table that shows the amount of water withdrawn in billion gallons per year, between 1950 and 2005,

Year 1950 1965 1975 1990 2005

Amount 180 340 420 410 410

Since, 420 > 180,

Also, 50 percent of water withdrawn is used up by thermometric power plants.

Thus, thermal power plants in 1975 > thermal power plant in 1950

Therefore, the number of thermal power plants increased between 1950 and 1975.

Second option is correct.

Note : first option is incorrect.

Since, by the above table,

The least amount of water used by power plant is 180 billion gallons per year,

Thus, third option is incorrect,

Now, the table does not represents energy produced,

Fourth option can not be said correct.

Mathematics

Aontario manufacturing company produces steel housings for electrical equipment. the main component part of the housing is a steel trough that is made out of a 14-gauge steel coil. it is produced using a 250-ton progressive punch press with a wipe-down operation that puts two 90-degree forms in the flat steel to make the trough. the distance from one side of the form to the other is critical because of weatherproofing in outdoor applications. the company requires that the width of the trough be between 8.31 inches and 8.61 inches. the widths of

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P Answered by Master

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.

Aontario manufacturing company produces steel housings for electrical equipment. the main component

Mathematics

Aontario manufacturing company produces steel housings for electrical equipment. the main component part of the housing is a steel trough that is made out of a 14-gauge steel coil. it is produced using a 250-ton progressive punch press with a wipe-down operation that puts two 90-degree forms in the flat steel to make the trough. the distance from one side of the form to the other is critical because of weatherproofing in outdoor applications. the company requires that the width of the trough be between 8.31 inches and 8.61 inches. the widths of

Step-by-step answer

P Answered by Master

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.

Business

Suppose the following data were taken from the 2022 and 2021 financial statements of American Eagle Outfitters. (All numbers, including share data, are in thousands.) 2022 2021 Current assets $ 871,500 $972,000 Total assets 1,908,500 1,786,000 Current liabilities 415,000 360,000 Total liabilities 564,916 528,656 Net income 197,760 410,590 Net cash provided by operating activities 322,000 498,600 Capital expenditures 289,000 290,200 Dividends paid on common stock 82,000 126,700 Weighted-average common shares outstanding 206,000 216,100 (a) Calculate

Step-by-step answer

P Answered by PhD

4

Kindly check explanation

Explanation:

Given the following :

20222021

Current asset871,500___972,000

Total assets1,908,500__ 1,786,000

Current liabilities___415,000 360,000

Total liabilities 564,916 528,656

Net income197,760 410,590

Net cash (from OP)__322,000 498,600

Capital expenditures_289,000___ 290,200

Dividends paid(CS)___82,000 126,700

Weighted-average common shares outstanding 206,000 216,100

*(OP) = Operating activities

*(CS) = common stock

Current ratio for each year:

2022:

Current asset / current liability

$871,500 / $415,000 = 2.1 : 1

2021:

$972,000 / $360,00 = 2.7 : 1

EARNING PER SHARE :

Net income / weighted average shares outstanding

2022:

$197,760 / 206,000 = $0.96

2021:

$410,590 / 216,100 = $1.90

DEBT TO ASSET RATIO:

Total liabilities / Total asset

2022:

$564,916 / $1,908,500 = 0.296

2021:

$528,656 / 1,786,000 = 0.296

FREE CASH FLOW :

Net cash (from OP) - Capital expenditure - Dividend paid on common stock

2022:

$322,000 - $289,000 - $82,000 = - $49,000

2021:

$498,600 - $290,200 - $126,700 = $81,700

410 is what percent of 65?

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