program 41602204_258983_UL×2.2_Vetri_12v_UQM_Analog_r189_LOS in process_20Jun2023.xcal

Screen 1

Name: C_USM_flm_DosingRate_Tbl

\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
\multicolumn{2}{|c|}{C_USM_flm_DosingRate_ты} & \multicolumn{2}{|c|}{[tex]$X$[/tex]: C\_USM\_PC\_DosingRate} & \multicolumn{2}{|c|}{[tex]$Y$[/tex]: C\_USM\_pro\_DosingRate} & \multirow{2}{}{\begin{tabular}{l}
Unit [tex]$1 ml / sec$[/tex] \\
850.0
\end{tabular}} & \multicolumn{3}{|c|}{X-Axis Unit: \% [tex]$\quad$[/tex] Y-Axis Unit: KPa} & \multirow{2}{
}{1050.0} & \multirow{2}{}{1100.0} & \multirow{2}{}{1200.0} \\
\hline
[tex]$X / Y$[/tex] & 600.0 & 650.0 & 700.0 & 750.0 & 800.0 & & 900.0 & 950.0 & 1000.0 & & & \\
\hline
0.000 & 0.0000 & 0.0000 & 0.0000 & 0.0000 & 0.0000 & 0.0000 & 0.0000 & 0.0000 & 0.0000 & 0.0000 & 0.0000 & 0.0000 \\
\hline
5.000 & 0.0532 & 0.0557 & 0.0576 & 0.0596 & 0.0615 & 0.0635 & 0.0654 & 0.0669 & 0.0688 & 0.0703 & 0.0723 & 0.0752 \\
\hline
10.000 & 0.1064 & 0.1108 & 0.1152 & 0.1191 & 0.1230 & 0.1270 & 0.1304 & 0.1338 & 0.1377 & 0.1411 & 0.1440 & 0.1509 \\
\hline
20.000 & 0.2129 & 0.2217 & 0.2300 & 0.2383 & 0.2461 & 0.2534 & 0.2607 & 0.2681 & 0.2749 & 0.2817 & 0.2886 & 0.3013 \\
\hline
30.000 & 0.3193 & 0.3325 & 0.3452 & 0.3574 & 0.3691 & 0.3804 & 0.3916 & 0.4019 & 0.4126 & 0.4229 & 0.4326 & 0.4521 \\
\hline
40.000 & 0.4263 & 0.4434 & 0.4604 & 0.4766 & 0.4922 & 0.5073 & 0.5220 & 0.5361 & 0.5503 & 0.5635 & 0.5767 & 0.6025 \\
\hline
50.000 & 0.5327 & 0.5542 & 0.5752 & 0.5952 & 0.6147 & 0.6338 & 0.6523 & 0.6699 & 0.6875 & 0.7046 & 0.7212 & 0.7534 \\
\hline
60.000 & 0.6392 & 0.6650 & 0.6904 & 0.7144 & 0.7378 & 0.7607 & 0.7827 & 0.8042 & 0.8252 & 0.8452 & 0.8652 & 0.9038 \\
\hline
70.000 & 0.7456 & 0.7759 & 0.8052 & 0.8335 & 0.8608 & 0.8877 & 0.9131 & 0.9380 & 0.9624 & 0.9863 & 1.0098 & 1.0547 \\
\hline
80.000 & 0.8521 & 0.8867 & 0.9204 & 0.9526 & 0.9839 & 1.0142 & 1.0435 & 1.0723 & 1.1001 & 1.1274 & 1.1538 & 1.2051 \\
\hline
90.000 & 0.9585 & 0.9976 & 1.0356 & 1.0718 & 1.1069 & 1.1411 & 1.1743 & 1.2061 & 1.2378 & 1.2681 & 1.2979 & 1.3560 \\
\hline
97.000 & 1.0332 & 1.0752 & 1.1162 & 1.1553 & 1.1929 & 1.2300 & 1.2656 & 1.3003 & 1.3340 & 1.3667 & 1.3989 & 1.4614 \\
\hline
\end{tabular}



Answer :

Alright class, let's solve this problem step by step.

### Step-by-Step Solution

1. Understanding the Problem:
Olivia starts with [tex]$23. She then buys five bagels. Each bagel costs $[/tex]3. We need to determine two things:
- The total amount of money Olivia spent.
- How much money Olivia is left with after buying the bagels.

2. Analyzing the Costs:
- The cost of one bagel is [tex]$3. - Olivia buys five bagels. 3. Calculating the Total Spent: - To find out how much Olivia spent in total, we multiply the cost of one bagel by the number of bagels. \[ \text{Total Cost} = \text{Cost per Bagel} \times \text{Number of Bagels} \] Substituting the given values: \[ \text{Total Cost} = 3 \times 5 = 15 \] Therefore, Olivia spends $[/tex]15 on bagels.

4. Calculating the Remaining Money:
- To find out how much money Olivia still has after buying the bagels, we subtract the total amount spent from her initial amount of money.
[tex]\[ \text{Remaining Money} = \text{Initial Money} - \text{Total Cost} \][/tex]

Substituting the values:
[tex]\[ \text{Remaining Money} = 23 - 15 = 8 \][/tex]

5. Conclusion:
- The total amount of money that Olivia spends on bagels is [tex]$15. - The amount of money Olivia has left after buying the bagels is $[/tex]8.

So, the detailed answer to the question is:
- Olivia spent [tex]$15 on bagels. - Olivia has $[/tex]8 left after the purchase.