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strojnjashka [21]
3 years ago
5

A catering company prepared and served 300 meals at an anniversary celebration last week using eight workers. The week before, s

ix workers prepared and served 240 meals at a wedding reception. a. For which event was the labor productivity higher
Mathematics
1 answer:
swat323 years ago
3 0

Answer:

II case.

Step-by-step explanation:

Given that a catering company prepared and served 300 meals at an anniversary celebration last week using eight workers.

The week before, six workers prepared and served 240 meals at a wedding reception.

Productivity is normally measured by number of outputs/number of inputs

Here we can measure productivity as

no of meals served/no of workers

In the I case productivity =\frac{300}{8} \\=37.5

In the II case productivity = \frac{240}{6} \\=40

Obviously II case productivity is more as per worker 40 meals were served which is more than 37.5 meals per worker in the I case.

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Answer:

1

Step-by-step explanation:

First, convert all the secants and cosecants to cosine and sine, respectively. Recall that csc(x)=1/sin(x) and sec(x)=1/cos(x).

Thus:

\frac{sec(x)}{cos(x)} -\frac{sin(x)}{csc(x)cos^2(x)}

=\frac{\frac{1}{cos(x)} }{cos(x)} -\frac{sin(x)}{\frac{1}{sin(x)}cos^2(x) }

Let's do the first part first: (Recall how to divide fractions)

\frac{\frac{1}{cos(x)} }{cos(x)}=\frac{1}{cos(x)} \cdot \frac{1}{cos(x)}=\frac{1}{cos^2(x)}

For the second term:

\frac{sin(x)}{\frac{cos^2(x)}{sin(x)} } =\frac{sin(x)}{1} \cdot\frac{sin(x)}{cos^2(x)}=\frac{sin^2(x)}{cos^2(x)}

So, all together: (same denominator; combine terms)

\frac{1}{cos^2(x)}-\frac{sin^2(x)}{cos^2(x)}=\frac{1-sin^2(x)}{cos^2(x)}

Note the numerator; it can be derived from the Pythagorean Identity:

sin^2(x)+cos^2(x)=1; cos^2(x)=1-sin^2(x)

Thus, we can substitute the numerator:

\frac{1-sin^2(x)}{cos^2(x)}=\frac{cos^2(x)}{cos^2(x)}=1

Everything simplifies to 1.

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