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Ray Of Light [21]
3 years ago
5

Please help me answer this, I don't get what I'm supposed to do TvT

Mathematics
1 answer:
kompoz [17]3 years ago
6 0

Answer:

You save 86 cents per pound.

Step-by-step explanation:

According to the given problem, grocery store B charges chicken 5 pounds of chicken for $38.20. Consider finding the unit price:

To find the unit price of any item, do the following calculation:

Total Price / Total amount = $$$ per amount.

The unit price of store b is \frac{38.20}{5}=7.64 . In other words, every pound of chicken costs $7.64 at store b.

For store a, we are provided with a table. Given how the question is being asked, we should <u>expect a higher unit price</u>. We can take any charge to find the unit price since the price should be consistent no matter how many pounds you buy. I will calculate the first row:

\frac{12.75}{1.5}=8.5

In other words, every pound of chicken costs $8.50 at store a. <u>This price is higher</u>. You can verify this is the correct unit price by multiplying the unit price with any amount of pounds provided at the table. You should get the total cost.

So, now that we have both unit prices, we can calculate the difference to find out how much we save per pound when choosing store b:

8.50-7.64=0.86.

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Show work explain with formulas.
Sedaia [141]

Answer:

\large\boxed{1.\ a_{44}=481,\ \sum\limits_{n=1}^\infty(11n-3),\ \text{the sum not exist}}

\large\boxed{3.\ \sum\limits_{n=1}^{47}(9n+16)=10,904}

\large\boxed{4.\ \sum\limits_{n=1}^\infty\bigg(6(2^n)\bigg),\ \text{the sum not exist}}

Step-by-step explanation:

1.\\8+19+30+41+...\\\\19-8=11\\30-19=11\\41-30=11\\\\\text{It's an arithmetic series.}\ a_1=8,\ d=11.\\\\\text{The formula for the n-th term of an arithmetic sequence:}\\\\a_n=a_1+(n-1)d\\\\\text{Substitute:}\\\\a_n=8+(n-1)(11)\qquad\text{use the distributive property}\ a(b-c)=ab-ac\\a_n=8+11n-11\\\boxed{a_n=11n-3}\\\\\text{Calculate the 44th term. Put n = 44 to the formula:}\\\\a_{44}=(11)(44)-3=484-3=481

\text{The summation notation:}\\\\\sum\limits_{n=1}^\infty(11n-3)\\\\\text{The sum not exist, because}\ d>1,\ \text{therefore}\ 11n-3\rightarrow\infty.

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

3.\\25+34+43+52+...+436\\\\34-25=9\\43-34=9\\52-43=9\\\\\text{It's an arithmetic series.}\ a_1=25,\ d=9.\\\\\text{The formula for the n-th term of an arithmetic sequence:}\\\\a_n=a_1+(n-1)d\\\\\text{Substitute:}\\\\a_n=25+(n-1)(9)\\a_n=25+9n-9\\a_n=9n+16\\\\\text{Calculate which the term of arithmetic series is the number 439.}\\\text{Put}\ a_n=439:\\\\9n+16=439\qquad\text{subtract 16 from both sides}\\9n=423\qquad\text{divide both sides by 9}\\n=47

\text{Therefore we have the series in summation notation:}\\\\\sum\limits_{n=1}^{47}(9n+16)\\\\\text{For calculation of the sum we use the formula of a sum of terms}\\\text{of an arithmetic sequence:}\\\\S_n=\dfrac{a_1+a_n}{2}\cdot n\\\\\text{Substitute}\ n=47,\ a_1=25,\ a_{47}=439:\\\\S_{47}=\dfrac{25+439}{2}\cdot47=\dfrac{464}{2}\cdot47=232\cdot47=10904\\\\\sum\limits_{n=1}^{47}(9n+16)=10,904

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

4.\\12+24+48+...\\\\24:12=2\\48:24=2\\\\\text{It's\ a\ geometric series}\ a_1=12,\ r=2.\\\\\text{The formula for the n-th term of a geometic sequence:}\\\\a_n=a_1r^{n-1}\\\\\text{Substitute:}\\\\a_n=(12)(2^{n-1})\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\a_n=(12)\left(\dfrac{2^n}{2^1}\right)\\\\a_n=(6)(2^n)\\\\\text{The summation notation:}\\\\\sum\limits_{n=1}^\infty\bigg(6(2^n)\bigg)\\\\\text{The sum not exist, because}\ r>1,\ \text{therefore}\ 6(2^n)\to\infty

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shusha [124]

Answer:

11

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Step-by-step explanation:

3 hrs to travel 45 miles downstream

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