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enot [183]
3 years ago
13

Decide whether the following statement makes sense (or is clearly true) or does not make sense (or is clearly false).Explain you

r reasoning.
Emma and Emily are good friends who do everything together, spending the same amount on eating out, entertainment, and other leisure activities. Yet Emma has a negative monthly cash flow while Emily's is positive, because Emily has more income.

Choose the correct answer below.

A.This statement makes sense because Emily makes more money than Emma, so she has a greater cash flow at the end of the month.

B.The statement makes sense because even though Emma and Emily spend the same amount on entertainment expenses, they may spend different amounts on other expenses. Regardless of income, Emma's and Emily's cash flows could be positive or negative.

C.This statement does not make sense because Emma uses a credit card for her expenses and Emily uses cash for her expenses.

D.This statement does not make sense because Emily's rent could be lower than Emma's, which means that she will have a greater monthly cash flow.
Mathematics
1 answer:
kondaur [170]3 years ago
7 0

Answer: A.This statement makes sense because Emily makes more money than Emma, so she has a greater cash flow at the end of the month.

Step-by-step explanation:

Emily has more monthly income and even if the duo spend exactly the same figure on everything they participate in, Emma will always come off short (negative monthly cash flow) because she does not earn as much as Emily does but spends equally.

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How to solve part ii and iii
iragen [17]

(i) Given that

\tan^{-1}(x) + \tan^{-1}(y) + \tan^{-1}(xy) = \dfrac{7\pi}{12}

when x=1 this reduces to

\tan^{-1}(1) + 2 \tan^{-1}(y) = \dfrac{7\pi}{12}

\dfrac\pi4 + 2 \tan^{-1}(y) = \dfrac{7\pi}{12}

2 \tan^{-1}(y) = \dfrac\pi3

\tan^{-1}(y) = \dfrac\pi6

\tan\left(\tan^{-1}(y)\right) = \tan\left(\dfrac\pi6\right)

\implies \boxed{y = \dfrac1{\sqrt3}}

(ii) Differentiate \tan^{-1}(xy) implicitly with respect to x. By the chain and product rules,

\dfrac d{dx} \tan^{-1}(xy) = \dfrac1{1+(xy)^2} \times \dfrac d{dx}xy = \boxed{\dfrac{y + x\frac{dy}{dx}}{1 + x^2y^2}}

(iii) Differentiating both sides of the given equation leads to

\dfrac1{1+x^2} + \dfrac1{1+y^2} \dfrac{dy}{dx} + \dfrac{y + x\frac{dy}{dx}}{1+x^2y^2} = 0

where we use the result from (ii) for the derivative of \tan^{-1}(xy).

Solve for \frac{dy}{dx} :

\dfrac1{1+x^2} + \left(\dfrac1{1+y^2} + \dfrac x{1+x^2y^2}\right) \dfrac{dy}{dx} + \dfrac y{1+x^2y^2} = 0

\left(\dfrac1{1+y^2} + \dfrac x{1+x^2y^2}\right) \dfrac{dy}{dx} = -\left(\dfrac1{1+x^2} + \dfrac y{1+x^2y^2}\right)

\dfrac{1+x^2y^2 + x(1+y^2)}{(1+y^2)(1+x^2y^2)} \dfrac{dy}{dx} = - \dfrac{1+x^2y^2 + y(1+x^2)}{(1+x^2)(1+x^2y^2)}

\implies \dfrac{dy}{dx} = - \dfrac{(1 + x^2y^2 + y + x^2y) (1 + y^2) (1 + x^2y^2)}{(1 + x^2y^2 + x + xy^2) (1+x^2) (1+x^2y^2)}

\implies \dfrac{dy}{dx} = -\dfrac{(1 + x^2y^2 + y + x^2y) (1 + y^2)}{(1 + x^2y^2 + x + xy^2) (1+x^2)}

From part (i), we have x=1 and y=\frac1{\sqrt3}, and substituting these leads to

\dfrac{dy}{dx} = -\dfrac{\left(1 + \frac13 + \frac1{\sqrt3} + \frac1{\sqrt3}\right) \left(1 + \frac13\right)}{\left(1 + \frac13 + 1 + \frac13\right) \left(1 + 1\right)}

\dfrac{dy}{dx} = -\dfrac{\left(\frac43 + \frac2{\sqrt3}\right) \times \frac43}{\frac83 \times 2}

\dfrac{dy}{dx} = -\dfrac13 - \dfrac1{2\sqrt3}

as required.

3 0
2 years ago
To find how high school students feel about hot lunches, Adam walks to the nearest high school and gives a survey postcard to ev
Eddi Din [679]
The surveyor selects the twentieth student who enters the building. The student do not enter the building to be surveyed but for other reasons and the researcher take advantage of that and selects them. This is a form of continence sampling.
 The researcher selects every twentieth student and not any other. Therefore, the sampling is systematic in nature.
Finally, students selected to participate in the survey are free to choose if to participate in he survey or not. Therefore, it is voluntary kind of sampling.

Therefore, the applicable sampling methods are convenience, systematic, and voluntary.
3 0
3 years ago
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The manager of a fast food restaurant is supposed to serve coffee that is 160 degrees Fahrenheit. A random sample of 20 cups of
evablogger [386]

Answer:

C. No, the Normal/large sample condition is not met.

Step-by-step explanation:

There is only 20 cup and for a sample to meet the Normal/large sample condition it must be greater than 30 samples or have a normal distribution or have no skewness or outliers  

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3 years ago
A store had 4 packs of paper for $1.64. How much would it cost if you were to buy 7 packs ??
Shalnov [3]

Answer: $2.87

Step-by-step explanation:

0.41 per pack

0.41 x 7 = 2.87

4 0
3 years ago
SkinnieQueen2019 got a 23/25 on her math test. What percent did she get?
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92% 23 divided by 25 is 0.92
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3 years ago
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