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svet-max [94.6K]
3 years ago
13

A boat ride costs $6.50 for one adult and $2.50 for one child. Which expression represents the total price a group of x adults a

nd y children would pay to ride the boat?
Mathematics
2 answers:
Tpy6a [65]3 years ago
3 0
6.5x + 2.5y is the correct answer
vovangra [49]3 years ago
3 0
6.50x+2.50y 6.50 times (x) the amount of adults riding 2.50 times (y) the amount of kids riding
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During the first 3 days of the week, Mike read an average of 28 pages per day. During the next 4 days, Mike averaged 42 pages pe
Anestetic [448]

Answer:

36 pages per day on average

Step-by-step explanation:

Calculate the average.

3 * 28 (The first 3 days of the week where Mike reads 28 pages per day) = 84

4 * 42 (The next 4 days where Mike reads 42 pages per day ) = 168

168 + 84 = 252



To find the average using this total, divide it by the total amount of days of reading (seven days in this problem).

252 / 7 = 36

6 0
2 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

b) For this case we are 98% confidence that the true deviation for the population of interest is between 19.08 and 66.56

Step-by-step explanation:

423.6, 487.3, 453.2, 402.9, 483.0, 477.7, 442.3, 418.4, 459.0

Part a

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:

s=sqrt{\frac{\sum_{i=1}^8 (x_i -\bar x)^2}{n-1}}
And in order to find the sample mean we just need to use this formula:
[tex]\bar x =\frac{\sum_{i=1}^n x_i}{n}

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:

df=n-1=9-1=8

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

\chi^2_{1- \alpha/2}=1.65

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

For this case we are 98% confidence that the true deviation for the population of interest is between 19.08 and 66.56

4 0
3 years ago
7k÷3=21. What is the value of k?
viva [34]
7k/3 = 21
x 3       x3
7k = 63
/7      /7
k = 9
(7(9)/3 = 21    =    63/3 = 21    =    21 = 21)
5 0
3 years ago
Read 2 more answers
What are the partial products of 34 and 57
Brut [27]
<span>2257987 is the answer</span>
5 0
3 years ago
The equation y = 5.50x represents the cost y (in dollars) for x visits to the
mr Goodwill [35]

In Graph x-axis represents number of visits and y-axis represents cost.

As graph comes up to be a linear one, so we can clearly say that cost is increasing linearly in multiples of 5.5 with increase in number of visits.

Example :

If museum is visited once then cost (y) = 5.5 x 1 = 5.5

If museum is visited twice then cost (y) = 5.5 x 2 = 11

If museum is visited thrice then cost (y) = 5.5 x 3 = 16.5

... cost (y) goes on in creasing when number of visits (x) increase with multiples of 5.5

5 0
3 years ago
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