What is the amount earned by a car salesman if he sold a truck for $37.500 at a 2% commission? 15​

Answer:

$2,625 ( i think )

If im right add me on snap pls

idk_marqo

The range of the starting salaries is 53,000

Given,

The box and whisker plot below shows the starting salaries

The number of graduates in a small college = 120

We have to find the range of the starting salaries;

Range of a data;

The difference between the highest and lowest values for a given data collection is the range in statistics. For instance, the range will be 10 - 2 = 8 if the given data set is 2, 5, 8, 10, and 3. As a result, the range may alternatively be thought of as the distance between the highest and lowest observation.

Here,

Lowest value = 19,000

Highest value = 72,000

Then,

Range of the set = 72,000 - 19,000 = 53,000

That is,

The range of the given data set is 53,000

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To find the monthly interest rate, simply divide the APR by 12. I think the monthly interest rate is 1.4%

Answer:

5.

Step-by-step explanation:

15.90 / 3.18 =

Answer:

The Area of the trapezium is \(60cm^2\)

Step-by-step explanation:

Given,

Height (h)= 8cm

Sum of the parallel sides (a+b)= 15cm

Area of trapezium = \(1/2*h*(a+b)\)

                              =\(1/2*8*15\)

                              =\(4*15\)

Area of trapezium=60cm^2

Answer: Area = 60 cm²

Step-by-step explanation:

To find the area of a trapezium we use the formula

Area of trapezium = 1/2 (sum of parallel sides) × height

∴ Area = 1/2 ×15 cm × 8 cm

           = 60 cm²

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Answer:

Step-by-step explanation

Okay so lets solve step by step/

15% can be written as a  fraction: 15/100

The of in the equation means multiplying.

15/100= 3/20

3/20*9= 270/20

27/2 is answer

The weight of the man, 5000 miles above the center of the earth is 48 lbs.

What is Division method?

Division method is used to distributing a group of things into equal parts. Division is just opposite of multiplications.

For example, dividing 20 by 2 means splitting 20 into 2 equal groups of 10.

We know that;

The joint variation that is given in the problem above can be expressed as follows:

⇒ w = k/d²

where, w is the weight, k is the variation constant, and d is the distance from the center of the earth.

Now, Substituting the first scenario to obtain the value of k,

150 = k/(4000)²

k = 2.4 x 10⁹

Using this value for the second scenario,

⇒ w = (2.4 x 10^9) / (5,000)²

       = 48 lbs

Thus, The weight of the man, 5000 miles above the center of the earth is 48 lbs.

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Answer:

101.25

Step-by-step explanation:

75.00 x 35% = 26.25

26.25 + 75.00 = 101.25

Answer:

you plug -6 and get 8x(1-(-6))=8×7=56

Step-by-step explanation:

Answer:

56

Step-by-step explanation:

f(-6) = 8(1 - (-6))

f(-6) = 8(1 + 6)

f(-6) = 8(7)

g(-6) = 56

Best of Luck!

Answer: Statistic.

Step-by-step explanation: Statistic are used to refer to numerical measurement pertaining to a sample. A sample refers to a subset of members chosen from a population, which means a smaller representation of the larger group. Hence, in the scenario above, the population is the entire seniors which makes up the group. While the sample is the subset of seniors drawn from the whole. Hence, statistical measures calculated or derived from the sample is called statistic. Hence, the 25% obtained is a statistic.

The probability of the four events are: Event 1: 1/84Event 2: 3/14Event 3: 15/28 Event 4: 5/21

The total number of drinks = 9The number of sweet teas = 3The number of unsweetened teas = 6If you select 3 drinks at random, the following events can take place:

Event 1: All three drinks are sweet teas. The probability of event 1 = (Number of ways in which all three drinks can be sweet teas) / (Number of ways to select 3 drinks)The number of ways in which all three drinks can be sweet teas = 3C3 = 1 (because all three sweet teas are already fixed)The number of ways to select 3 drinks = 9C3 = (9 × 8 × 7)/(3 × 2 × 1) = 84Therefore, the probability of event 1 = 1/84 = 1/84

Event 2: Exactly two drinks are sweet teas. The probability of event 2 = (Number of ways in which two drinks are sweet teas and one is an unsweetened tea) / (Number of ways to select 3 drinks)The number of ways in which two drinks are sweet teas and one is an unsweetened tea = (3C2 × 6C1) = 18 (because you can choose 2 sweet teas from 3 and 1 unsweetened tea from 6)The number of ways to select 3 drinks = 9C3 = (9 × 8 × 7)/(3 × 2 × 1) = 84Therefore, the probability of event 2 = 18/84 = 3/14

Event 3: Exactly one drink is a sweet tea. The probability of event 3 = (Number of ways in which one drink is a sweet tea and the other two are unsweetened teas) / (Number of ways to select 3 drinks)The number of ways in which one drink is a sweet tea and the other two are unsweetened teas = (3C1 × 6C2) = 45 (because you can choose 1 sweet tea from 3 and 2 unsweetened teas from 6)The number of ways to select 3 drinks = 9C3 = (9 × 8 × 7)/(3 × 2 × 1) = 84Therefore, the probability of event 3 = 45/84 = 15/28

Event 4: All three drinks are unsweetened teas. The probability of event 4 = (Number of ways in which all three drinks can be unsweetened teas) / (Number of ways to select 3 drinks)The number of ways in which all three drinks can be unsweetened teas = 6C3 = 20 (because you can choose 3 unsweetened teas from 6)The number of ways to select 3 drinks = 9C3 = (9 × 8 × 7)/(3 × 2 × 1) = 84 Therefore, the probability of event 4 = 20/84 = 5/21

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Rounded to the nearest tenth, the acceleration in inches per second squared is approximately 15222.8 in/s².

To convert the acceleration from meters per second squared (m/s²) to inches per second squared (in/s²), we need to use the conversion factor between the two units.

1 meter is equal to 39.37 inches.

To convert the units, we can set up the following conversion factor:

1 m/s² = (39.37 in/m)^2 = 1550.0031 in/s²

Now, we can multiply the given acceleration in m/s² by the conversion factor to obtain the acceleration in in/s²:

Acceleration in in/s² = 9.81 m/s² * 1550.0031 in/s²

Acceleration in in/s² ≈ 15222.7568 in/s²

Rounded to the nearest tenth, the acceleration in inches per second squared is approximately 15222.8 in/s².

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Answer:

Sue rode 1885 miles total

Step-by-step explanation:

There are 31 days in March, 30 in April, and 30 in May.

31 * 12 + 30 * 12 + 30 * 12 = 1092

There are 30 days in June and 31 in august.

30 * 13 + 31 * 13 = 793

Now we find the total:

1092 + 793 = 1885

Hi!

Now I can answer.

I used systems of equations That is how I solved it. Always remember to check your answer by plugging it back into the word problem. If you aren't comfortable with systems of equations then find some help in this area.

(.85x)1.25=y new line

x + 10=y new line

x=y-10 new line

(.85(y-10))1.25=y new line

(.85y-8.5)1.25=y new line

1.0625y-10.625=y new line

.0625y=10.625 new line

y=170 new line

 x=160

Hope this helps!!

~~PicklePoppers~~

\(x^2-8x+10=x^2-8x+16-6=(x-4)^2-6\)

The ramp has a volume capacity of around 37.5 cubic feet.

How to solve

Calculating the volume of a triangular prism shaped ramp follows a specific formula. First, identify the shape and refer to the equation:

Volume = (1/2) * Base Area * Height

To begin, the base is always in a right-angled triangle form, where its area depends on its width denoted as W, and height noted specifically as H.

Based on these computations, knowing that its triangular base measures 5 feet at its length while taking its height equates 3 feet therefore assessing how much it covers can be found using:

Triangle Area = (1/2) * Base * Height

Substituting values into solving areas leads us to have an answer of over 7 square feet.

Next is computing the volume of the said triangular prism. Placing all given data results in the same formula:

Volume = (1/2) * Base Area * Height

With one-half of the product of both areas which resulted in 7.5 square feet and then multiplying the figure by 10 ft, this yields us an approximate volume of 37.5 cubic feet.

In conclusion, we have determined that the ramp has a volume capacity of around 37.5 cubic feet.

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Length (L) = 10 feet

Width (W) = 5 feet

Height (H) = 3 feet

Mr. González designed a ramp for his car with a length of 10 feet, a width of 5 feet, and a height of 3 feet. What is the volume of the ramp in cubic feet?

I’m not sure if this is correct but i think 540 because when u divide 884/221 u get 4 then multiply that by 135.

Answer:

=−129.25224x

Step-by-step explanation:

(−9.206x)(14.04)  you just simplify

(-9.206) x (14.04) =

Answer: -129.25224

Answer:

124.5252844501

Step-by-step explanation:

to estimate we have 125

Answer:

50.75

Step-by-step explanation:

We have:

\(E[g(x)] = \int\limits^{\infty}_{-\infty} {g(x)f(x)} \, dx \\\\= \int\limits^{1}_{-\infty} {g(x)(0)} \, dx+\int\limits^{6}_{1} {g(x)\frac{2}{x} } \, dx+\int\limits^{\infty}_{6} {g(x)(0)} \, dx\\\\= \int\limits^{6}_{1} {g(x)\frac{2}{x} } \, dx\\\\=\int\limits^{6}_{1} {(4x+3)\frac{2}{x} } \, dx\\\\=\int\limits^{6}_{1} {(4x)\frac{2}{x} } \, dx + \int\limits^{6}_{1} {(3)\frac{2}{x} } \, dx\\\\=\int\limits^{6}_{1} {8} \, dx + \int\limits^{6}_{1} {\frac{6}{x} } \, dx\\\\\)

\(=8\int\limits^{6}_{1} \, dx + 6\int\limits^{6}_{1} {\frac{1}{x} } \, dx\\\\= 8[x]^{^6}_{_1} + 6 [ln(x)]^{^6}_{_1}\\\\= 8[6-1] + 6[ln(6) - ln(1)]\\\\= 8(5) + 6(ln(6))\\\\= 40 + 10.75\\\\= 50.74\)

Answer:

- sqaure root 7-3, sqaure root 7-3

Step-by-step explanation:

See Image below:)

Answer:

1/2 - 1/3,

1 - 1/2,

1/4 + 1/2

Step-by-step explanation:

1/4 + 1/2 = 1/4 + 2/4 = 3/4

1/2 - 1/3 = 3/6 - 2/6 = 1/6

1 - 1/2 = 2/2 - 1/2 = 1/2

Hence, the desired order is

1/2 - 1/3,

1 - 1/2,

1/4 + 1/2

A. The function h(x) = \(4log_{2} (x ) -2\) in its simplest form.

B. There is no solution to the system of nonlinear equations.

In order to obtain the solution, composite functions concept has been used.

What are composite functions?

Composite functions are those functions which are composed of other functions, with the input of one serving as the output of the other.

We are given two functions as

\(f(x) = log_{2} (x) +2\)

\(g(x) = log_{2} (x^{3} ) -4\)

A. We need to solve for h(x)

We are given that h(x) = f (x) + g(x)

So, on adding the above equations, we get

h(x) = \(log_{2} (x) +2+log_{2} (x^{3} ) -4\)

On combining the like terms, we get

h(x) = \(log_{2} (x) +log_{2} (x^{3} ) -4 +2\)

h(x) = \(log_{2} (x) +log_{2} (x^{3} ) -2\)

Now, using the law of logarithm, we get

h(x) = \(log_{2} (x * x^{3} ) -2\)

h(x) = \(log_{2} (x^{4} ) -2\)

h(x) =  \(4log_{2} (x ) -2\)

Thus, the function h(x) = \(4log_{2} (x ) -2\) in its simplest form.

B. We need to find the solution to the system of nonlinear equations.

Since, the equations are not given, so there is no solution to the system of nonlinear equations.

Hence, the function h(x) = \(4log_{2} (x ) -2\) in its simplest form.

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Answer:

\(\displaystyle f'(1) = \frac{3}{2}\)

\(\displaystyle f''(1) = \frac{3}{4}\)

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

BracketsParenthesisExponentsMultiplicationDivisionAdditionSubtractionLeft to Right

Algebra II

Exponential Rule [Rewrite]:                                                                            \(\displaystyle b^{-m} = \frac{1}{b^m}\)Exponential Rule [Root Rewrite]:                                                                   \(\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}\)

Calculus

Derivatives

Derivative Notation

Basic Power Rule:

f(x) = cxⁿf’(x) = c·nxⁿ⁻¹

Derivative Property [Addition/Subtraction]: \(\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]\)

Derivative Rule [Product Rule]:                                                                             \(\displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)\)

Derivative Rule [Quotient Rule]:                                                                           \(\displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}\)

Step-by-step explanation:

Step 1: Define

f(x) = x√x

f'(1) is x = 1 for 1st derivative

f''(1) is x = 1 for 2nd derivative

Step 2: Differentiate

[1st Derivative] Product Rule:                                                                       \(\displaystyle f'(x) = \frac{d}{dx}[x]\sqrt{x} + x\frac{d}{dx}[\sqrt{x}]\)[1st Derivative] Rewrite [Exponential Rule - Root Rewrite]:                         \(\displaystyle f'(x) = \frac{d}{dx}[x]\sqrt{x} + x\frac{d}{dx}[x^{\frac{1}{2}}]\)[1st Derivative] Basic Power Rule:                                                                 \(\displaystyle f'(x) = (1 \cdot x^{1 - 1})\sqrt{x} + x(\frac{1}{2}x^{\frac{1}{2}-1})\)[1st Derivative] Simply Exponents:                                                               \(\displaystyle f'(x) = (1 \cdot x^0)\sqrt{x} + x(\frac{1}{2}x^{\frac{-1}{2}})\)[1st Derivative] Simplify:                                                                                 \(\displaystyle f'(x) = \sqrt{x} + x(\frac{1}{2}x^{\frac{-1}{2}})\)[1st Derivative] Rewrite [Exponential Rule - Rewrite]:                                 \(\displaystyle f'(x) = \sqrt{x} + x(\frac{1}{2x^{\frac{1}{2}}})\)[1st Derivative] Rewrite [Exponential Rule - Root Rewrite]:                         \(\displaystyle f'(x) = \sqrt{x} + x(\frac{1}{2\sqrt{x}})\)[1st Derivative] Multiply:                                                                                 \(\displaystyle f'(x) = \sqrt{x} + \frac{x}{2\sqrt{x}}\)[2nd Derivative] Rewrite [Exponential Rule - Root Rewrite]:                     \(\displaystyle f'(x) = x^{\frac{1}{2}} + \frac{x}{2x^{\frac{1}{2}}}\)[2nd Derivative] Basic Power Rule/Quotient Rule [Derivative Property]: \(\displaystyle f''(x) = \frac{1}{2}x^{\frac{1}{2} - 1} + \frac{\frac{d}{dx}[x](2x^{\frac{1}{2}}) - x\frac{d}{dx}[2x^{\frac{1}{2}}]}{(2x^{\frac{1}{2}})^2}\)[2nd Derivative] Simplify/Evaluate Exponents:                                           \(\displaystyle f''(x) = \frac{1}{2}x^{\frac{-1}{2}} + \frac{\frac{d}{dx}[x](2x^{\frac{1}{2}}) - x\frac{d}{dx}[2x^{\frac{1}{2}}]}{4x}\)[2nd Derivative] Rewrite [Exponential Rule - Rewrite]:                               \(\displaystyle f''(x) = \frac{1}{2x^{\frac{1}{2}}} + \frac{\frac{d}{dx}[x](2x^{\frac{1}{2}}) - x\frac{d}{dx}[2x^{\frac{1}{2}}]}{4x}\)[2nd Derivative] Basic Power Rule:                                                             \(\displaystyle f''(x) = \frac{1}{2x^{\frac{1}{2}}} + \frac{(1 \cdot x^{1 - 1})(2x^{\frac{1}{2}}) - x(\frac{1}{2} \cdot 2x^{\frac{1}{2} - 1})}{4x}\)[2nd Derivative] Simply Exponents:                                                             \(\displaystyle f''(x) = \frac{1}{2x^{\frac{1}{2}}} + \frac{(1 \cdot x^0)(2x^{\frac{1}{2}}) - x(\frac{1}{2} \cdot 2x^{\frac{-1}{2}})}{4x}\)[2nd Derivative] Simplify:                                                                             \(\displaystyle f''(x) = \frac{1}{2x^{\frac{1}{2}}} + \frac{2x^{\frac{1}{2}} - x(\frac{1}{2} \cdot 2x^{\frac{-1}{2}})}{4x}\)[2nd Derivative] Multiply:                                                                             \(\displaystyle f''(x) = \frac{1}{2x^{\frac{1}{2}}} + \frac{2x^{\frac{1}{2}} - x(x^{\frac{-1}{2}})}{4x}\)[2nd Derivative] Rewrite [Exponential Rule - Rewrite]:                               \(\displaystyle f''(x) = \frac{1}{2x^{\frac{1}{2}}} + \frac{2x^{\frac{1}{2}} - x(\frac{1}{x^{\frac{1}{2}}})}{4x}\)[2nd Derivative] Multiply:                                                                             \(\displaystyle f''(x) = \frac{1}{2x^{\frac{1}{2}}} + \frac{2x^{\frac{1}{2}} - \frac{x}{x^{\frac{1}{2}}}}{4x}\)[2nd Derivative] Simplify:                                                                             \(\displaystyle f''(x) = \frac{1}{2x^{\frac{1}{2}}} + \frac{x^{\frac{1}{2}}}{4x}\)[2nd Derivative] Rewrite [Exponential Rule  - Root Rewrite]:                     \(\displaystyle f''(x) = \frac{1}{2\sqrt{x}} + \frac{\sqrt{x}}{4x}\)

Step 3: Evaluate

[1st Derivative] Substitute in x:                                                                     \(\displaystyle f'(1) = \sqrt{1} + \frac{1}{2\sqrt{1}}\)[1st Derivative] Evaluate Roots:                                                                     \(\displaystyle f'(1) = 1 + \frac{1}{2(1)}\)[1st Derivative] Multiply:                                                                                 \(\displaystyle f'(1) = 1 + \frac{1}{2}\)[1st Derivative] Add:                                                                                       \(\displaystyle f'(1) = \frac{3}{2}\)[2nd Derivative] Substitute in x:                                                                   \(\displaystyle f''(1) = \frac{1}{2\sqrt{1}} + \frac{\sqrt{1}}{4(1)}\)[2nd Derivative] Evaluate Roots:                                                                 \(\displaystyle f''(1) = \frac{1}{2(1)} + \frac{1}{4(1)}\)[2nd Derivative] Multiply:                                                                             \(\displaystyle f''(1) = \frac{1}{2} + \frac{1}{4}\)[2nd Derivative] Add:                                                                                     \(\displaystyle f''(1) = \frac{3}{4}\)

240 soldiers were transferred to the other camp. Let's assume the number of soldiers transferred to the other camp is "x." we can solve for the unknown variable representing the number of soldiers transferred.

Initially, the food provision was planned for 800 soldiers for 35 days. This means that the total food supply is equal to 800 soldiers multiplied by 35 days, which gives us a total of 28,000 soldier-days of food.

After some soldiers were transferred, the remaining soldiers had enough food to last for 50 days. So the food supply for the remaining soldiers can be calculated as the product of the number of remaining soldiers and the number of days, which gives us (800 - x) * 50 soldier-days of food.

Since the amount of food remains the same, we can set up the following equation:

\(28,000 = (800 - x) * 50\)

Now, let's solve this equation to find the value of x, representing the number of soldiers transferred to the other camp:

\(28,000 = 50 * 800 - 50x28,000 = 40,000 - 50x50x = 40,000 - 28,00050x = 12,000x = 12,000 / 50x = 240\)

By setting up the equation based on the total food supply for the initial and remaining soldiers, we can solve for the unknown variable representing the number of soldiers transferred. In this case, 240 soldiers were transferred to the other camp.

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The journal entries to record the receipt of the application money and its allotment to share capital and Allotment Account are as follows:

Journal Entries:

Debit Cash Rs. 300,000

Credit Share Application Account Rs. 300,000

(To record the receipt of the money on application)

Debit Share Application Account Rs. 300,000

Credit Share Capital Rs. 150,000

Credit Share Allotment Account Rs. 150,000

(To record the transfer of the application money to the Share Capital and the Allotment accounts, respectively.)

What are journal entries?

Journal entries are the initial records kept of business transactions.

Journal entries are made as transactions occur on a daily basis.

Transaction Analysis:

Authorized capital = Rs. 1,000,000

The number of shares issued = 5,000

The par value = Rs. 100

The premium = Rs. 10 (Rs. 100 x 10%)

The issuance price per share = Rs. 110 (Rs. 100 + Rs. 10)

The expected amount to be received on application = Rs. 30 per share

The number of applications received = 10,000

Application money = Rs. 300,000

Excess Application = Rs. 150,000

Cash Rs. 300,000

Share Application Account Rs. 300,000

Share Application Account Rs. 300,000

Share Capital Rs. 150,000

Share Allotment Account Rs. 150,000

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Answer:100

Step-by-step explanation:

40+40 =80 +20=100

Answer:

Practical domain:  0 ≤ x ≤ 3.907Practical Range: 0 ≤ y ≤ 84 where y is an integer, so we have the set {0,1,2,...,83,84}

The 3.907 is approximate.

====================================

Explanation:

x = number of hours that elapse

y = f(x) = number of tokens

If we use a graphing tool like a TI84 or GeoGebra, then the approximate solution to -3(2)^(x+1) + 90 = 0 is roughly x = 3.907

At around 3.907 hours is when the number of tokens is y = 0. Therefore, this is the approximate upper limit for the domain. The lower limit is x = 0.

The domain spans from x = 0 to roughly x = 3.907, and we shorten that down to 0 ≤ x ≤ 3.907

------------

Plug in x = 0 to find y = 84. This is the largest value in the range.

The smallest value is y = 0.

The range spans from y = 0 to y = 84, so we get 0 ≤ y ≤ 84

Keep in mind y is the number of tokens. A fractional amount of tokens does not make sense, so we must have y be a whole number 1,2,3,...,83,84.

The x value can be fractional because 3.907 hours for instance is valid.

------------

Extra info:

The function is decreasing. It goes downhill when moving to the right.The points (0,84) and (1,78) and (2,66) and (3,42) are on this exponential curve.A point like (2,66) means x = 2 and y = 66. It indicates: "after 2 hours, they will have 66 tokens remaining".

A: The linear equations are x + y = 125 and x = y + 45.

B: Everyday Angela spends 40 minutes in playing golf.

C: No, it is not possible for Angela to spend 80 minutes playing soccer every day.

What is a linear equation?

A linear equation is one that has a degree of 1 as its maximum value. No variable in a linear equation, thus, has an exponent greater than 1. A linear equation's graph will always be a straight line.

Part A:

Let x be the number of minutes Angela plays soccer every day

Let y be the number of minutes Angela plays golf every day

According to the problem statement the linear equations are -

x + y = 125 (total time playing both sports is 125 minutes)

x = y + 45 (Angela plays 45 more minutes of soccer than golf)

Part B:

Substitute x = y + 45 into the first equation -

(y + 45) + y = 125

Use the addition operation -

2y + 45 = 125

2y = 80

y = 40

Angela spends 40 minutes playing golf every day.

Part C:

If Angela spends 80 minutes playing soccer every day, then

x = 80

y + 45 = 80 (using the equation x = y + 45)

y = 35

But this would mean that she plays golf for 35 minutes and soccer for 80 minutes, which adds up to a total of 115 minutes.

This is not possible since the problem states that Angela plays both sports for a total of 125 minutes every day.

Therefore, it is not possible for Angela to have spent 80 minutes playing soccer every day.

To learn more about linear equation from the given link

https://brainly.com/question/2226590

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Answer:

\(8.3125\)pounds

Step-by-step explanation:

We are given that

Density of ethyl acetate at 25 degrees =\(0.902g/ml\)

We have to find the number of pounds in 133 ounces weigh.

To find the number of pounds in 133 ounces we will use unitary method

By using unitary method

We know that

1 ounce=0.0625 pounds

133 ounce=\(133\times 0.0625\)pounds

133 ounce=\(8.3125\)pounds

Hence,8.3125 pounds would 133 ounces weigh.