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fenix001 [56]
3 years ago
13

John went to the deli on his lunch hour he bought 1.3 lbs of roast beef and 0.6 lbs of cheese both were on sale for $3.99 per po

und. To the nearest cent how much was Johns total at the deli
Mathematics
1 answer:
aleksley [76]3 years ago
4 0

Johns total at the deli is $ 7.58

<u>Solution:</u>

Given, John went to the deli on his lunch hour he bought 1.3 lbs of roast beef and 0.6 lbs of cheese both were on sale for $3.99 per pound.  

We have to find the nearest cent how much was Johns total at the deli

Now, we know that, total cost = cost of beef + cost of cheese

Total cost = amount of beef bought \times cost of beef per pound + amount of cheese bought \times cost of cheese per pound  

Total amount = 1.3 \times 3.99 + 0.6 \times 3.99

Total amount = 3.99(1.3 + 0.6) = 3.99 x 1.9

Total amount = 7.581

Hence, the total amount was $7.58

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The annual net income of a company for the period 2007–2011 could be approximated by P(t) = 1.6t2 − 11t + 44 billion dollars (2
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P'(t) = 3.2 t -11

And we can set equal this derivate to 0 in order to find the critical point and we got:

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P''(t) = 3.2 >0

So then w can conclude that the value of t = 3.4375 represent the minimum value for the function and we can replace in the original function and we got:

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Step-by-step explanation:

For this case we have the following function:

P(t) = 1.6 t^2 -11t +44

Where P represent the annual net income for the period 2007-2011 and 2 \leq t \leq 7

And t represent the time in years since the start of 2005

In order to find the lowet income we need to use the derivate, given by:

P'(t) = 3.2 t -11

And we can set equal this derivate to 0 in order to find the critical point and we got:

3.2 t-11= 0

t = \frac{11}{3.2}= 3.4375

And we can calculate the second derivate and we got:

P''(t) = 3.2 >0

So then w can conclude that the value of t = 3.4375 represent the minimum value for the function and we can replace in the original function and we got:

P(3.4375) = 1.6(3.4375)^2 - (11*3.4375) +44 = 25.094

So then the minimum annual income occurs at t = 3.43 (between 2008 and 2009) and the value is 25.094

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3 years ago
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