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Luda [366]
3 years ago
14

Greg rented a truck for one day there was a base fee of $16.99 and there was additional charge of $.89 for each mile driven Greg

had pay $207.45 when he returned the truck for how many miles did he drive the truck
Mathematics
1 answer:
julia-pushkina [17]3 years ago
7 0
Let m represent the number of mile Greg drove. His rental charge was
   207.45 = 16.99 + 0.89m
Subtract 16.99 and divide by 0.89
   190.46 = 0.89 m
   214 = m

Greg drove the truck 214 miles.
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If you eat 4 medium strawberries you get 48%of your daily reccomended amount of vitiman c what fraction of your daily amount of
Nady [450]
<span>4 medium strawberries→</span><span> 48% of your daily recommend amount of vitamin c
x </span><span>medium strawberries→100% </span><span>of your daily recommend amount of vitamin c
4:48=x:100⇒48x=100*4
48x=400
x=8.33
8.33-4=4.33
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6 0
3 years ago
You need to invest $1000 in a bank account and are give two options. The first option is to earn $50 every month you leave the m
garri49 [273]

Answer:

<h3><u>Option 1</u></h3>

Earn $50 every month.

  • Let x = number of months the money is left in the account
  • Let y = the amount in the account
  • Initial amount = $1,000

\implies y = 50x + 1000

This is a <u>linear function</u>.

<h3><u>Option 2</u></h3>

Earn 3% interest each month.

(Assuming the interest earned each month is <u>compounding interest</u>.)

  • Let x = number of months the money is left in the account
  • Let y = the amount in the account
  • Initial amount = $1,000

\implies y = 1000(1.03)^x

This is an <u>exponential function</u>.

<h3><u>Table of values</u></h3>

<u />

\large \begin{array}{| c | l | l |}\cline{1-3} & \multicolumn{2}{|c|}{\sf Account\:Balance} \\ \cline{1-3} & \sf Option\:1 & \sf Option\:2 \\\sf Month & \sf \$50\:per\:mth & \sf 3\%\:per\:mth \\\cline{1-3} 0 & \$1000 & \$1000 \\\cline{1-3} 1 & \$1050 & \$1030 \\\cline{1-3} 2 & \$1100 & \$1060.90 \\\cline{1-3} 3 & \$1150 & \$1092.73 \\\cline{1-3} 4 & \$1200 & \$1125.51 \\\cline{1-3} 5 & \$1250 & \$1159.27 \\\cline{1-3} 6 & \$1300 & \$1194.05 \\\cline{1-3} 7 & \$1350 & \$1229.87 \\\cline{1-3}\end{array}

From the table of values, it appears that <u>Account Option 1</u> is the best choice, as the accumulative growth of this account is higher than the other account option.

However, there will be a point in time when Account Option 2 starts accruing more than Account Option 2 each month.  To find this, graph the two functions and find the <u>point of intersection</u>.

From the attached graph, Account Option 1 accrues more until month 32.  From month 33, Account Option 2 accrues more in the account.

<h3><u>Conclusion</u></h3>

If the money is going to be invested for less than 33 months then Account Option 1 is the better choice.  However, if the money is going to be invested for 33 months or more, then Account Option 2 is the better choice.

3 0
1 year ago
There is a function p, wher p(X) represents the perceent of aldult in the united states who owned a x years after 2011. What doe
masya89 [10]

Given: p(x) represents the percent of adults in the US who owned a <em><u>widget</u></em>  

Given: x represents the number of years after 2011

Answer:

p(5)= 68 means:

        68% of adults owned a <em><u>widget</u></em>

        5 years after 2011


⇒    68% of adults owned a <em><u>widget</u></em> in 2016

6 0
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