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3 years ago

17 times the sum of a number, n, and 31 is 300. Write as an equation.

Mathematics

1 answer:

Illusion [34]3 years ago

3 0

Answer:

17(n+31)=300

Step-by-step explanation:

17 times the sum of a number, n and 31 is 17(n+31)

and then set that equal to 300

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3 years ago

Daily high temperatures in St. Louis for the last week were as follows: 95, 92, 93, 92, 95, 90, 90 (yesterday). a) The

hichkok12 [17]

Answer:

a) T = 91.7 degrees

b) T = 90 degrees

c) MAD = 1.9

d) MSE = 5.05

Step-by-step explanation:

Given:

- Daily high temperatures in St. Louis for the last week were as follows:

95, 92, 93, 92, 95, 90, 90

Find:

a) Forecast the high temperature today, using a 3-day moving average.

b) Forecast the high temperature today, using a 2-day moving average.

c) Calculate the mean absolute deviation based on a 2-day moving average, covering all days in which you can have a forecast and an actual temperature.

d) The mean squared error for the 2-day moving average

Solution:

- The set of 3 day moving average is as follows:

4. (95 + 92 + 93) ÷ 3 = 93.33⁰C

5. (92 + 93 + 92) ÷ 3 = 92.33⁰C

6. (93 + 92 + 95) ÷ 3 = 93.33⁰C

7. (92 + 95 + 90) ÷ 3 = 92.33⁰C

8. (95 + 90 + 90) ÷ 3 = 91.667⁰C

- Now use these points on excel sheet to forecast the temperature for today. The line of best fit is given:

T = 91.7 degrees

- The set of 2 day moving average is as follows:

3. (95 + 92) ÷ 2 = 93.5⁰C

4. (95 + 93) ÷ 2 = 92.5⁰C

5. (93 + 92) ÷ 2 = 92.5⁰C

6. (92 + 95) ÷ 2 = 93.5⁰C

7. (95 + 90) ÷ 2 = 92.5⁰C

8. (90 + 90) ÷ 2 = 90⁰C

- Now use these points on excel sheet to forecast the temperature for today. The line of best fit is given:

T = 90 degrees

Error Error^2

3. 93.5⁰C 0.5 0.25

4. 92.5⁰C 0.5 0.25

5. 92.5⁰C 2.5 6.25

6. 93.5⁰C 3.5 12.25

7. 92.5⁰C 2.5 6.25

8. 90⁰C

- The mean absolute deviation as follows:

MAD = Sum of all errors / 5

MAD = (0.5+0.5+2.5+3.5+2.5) / 5

MAD = 1.9

- The mean squared error deviation as follows:

MSE = Sum of all error^2 / 5

MSE = (0.25+0.25+6.25+12.25+6.25) / 5

MSE = 5.05

4 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.

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2 years ago

there are two couches against a wall. if one takes up 4/9 of the wall and the other takes up 1/5 of the wall, how much of the wa

Angelina_Jolie [31]

4/9+1/5=20/45+9/45=29/45 of the wall is covered up by two couches

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4 years ago

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