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NikAS [45]
2 years ago
6

After eating a meal at a restaurant, we decided to tip 25%, building a grand total of $87.50. What was the price before the tip?


Mathematics
1 answer:
CaHeK987 [17]2 years ago
3 0

Answer:

65.62

Step-by-step explanation:

Given:

After eating a meal at a restaurant, we decided to tip 25%, building a grand total of $87.50.

To find:

Price before tip

Solution:

87.50 × 25% = 21.875

~Round~: 21.875 to 21.88

87.50 - 21 .88 = 65.62

Thus the price before the tip is 65.62

Check Answer:

<em>Formula: Higher number - Lower number ÷ original number × 100</em>

<em>Solve: </em>

<em>87.50 - 65.62 = 21.88</em>

<em>21.88 ÷ 87.50=0.25005714285</em>

<em>0.25005714285 × 100 = 25.0057142857</em>

<em>Round - 25%</em>

 

<u><em>~lenvy~</em></u>

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The school that Lisa goes to is selling tickets to the annual talent show. On the first day of ticket sales the school sold 4 se
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Let:
x = cost of senior citizen ticket
y = cost of student ticket

4x + 5y = 102
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3 years ago
Flow meters are installed in urban sewer systems to measure the flows through the pipes. In dry weatherconditions (no rain) the
ziro4ka [17]

Answer:

a) \frac{(8)(30.23)^2}{20.09} \leq \sigma^2 \leq \frac{(8)(30.23)^2}{1.65}

363.90 \leq \sigma^2 \leq 4430.80

Now we just take square root on both sides of the interval and we got:

19.08 \leq \sigma \leq 66.56

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Step-by-step explanation:

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The confidence interval for the population variance is given by the following formula:

\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}

On this case we need to find the sample standard deviation with the following formula:

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The sample deviation for this case is s=30.23

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

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The Confidence interval is 0.98 or 98%, the value of \alpha=0.02 and \alpha/2 =0.01, and the critical values are:

\chi^2_{\alpha/2}=20.09

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And replacing into the formula for the interval we got:

\frac{(8)(30.23)^2}{20.09} \leq \sigma^2 \leq \frac{(8)(30.23)^2}{1.65}

363.90 \leq \sigma^2 \leq 4430.80

Now we just take square root on both sides of the interval and we got:

19.08 \leq \sigma \leq 66.56

Part b

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