I WILL MARK BRAINLIEST PLEASE ANSWER
1 answer:
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<u>Part (a)</u>
The variable y is the dependent variable and the variable x is the independent variable.
<u>Part (b)</u>
The cost of one ticket is $0.75. Therefore, the cost of 18 tickets will be:
dollars
Now, we know that Kendall spent her money only on ride tickets and fair admission and that she spent a total of $33.50.
Therefore, the price of the fair admission is: $33.50-$13.50=$20
If we use y to represent the total cost and x to represent the number of ride tickets, the linear equation that can be used to determine the cost for anyone who only pays for ride tickets and fair admission can be written as:
......Equation 1
<u>Part (c)</u>
The above equation is logical because, in general, the total cost of the rides will depend upon the number of ride tickets bought and that will be 0.75x. Now, even if one does not take any rides, that is when x=0, they still will have to pay for the fair admission, and thus their total cost, y=$20.
Likewise, any "additional" cost will depend upon the number of ride tickets bought as already suggested. Thus, the total cost will be the sum of the total ride ticket cost and the fixed fair admission cost. Thus, the above Equation 1 is the correct representative linear equation of the question given.
Answer:
Step-by-step explanation:
Hello!
X: number of absences per tutorial per student over the past 5 years(percentage)
X≈N(μ;σ²)
You have to construct a 90% to estimate the population mean of the percentage of absences per tutorial of the students over the past 5 years.
The formula for the CI is:
X[bar] ± *
⇒ The population standard deviation is unknown and since the distribution is approximate, I'll use the estimation of the standard deviation in place of the population parameter.
Number of Absences 13.9 16.4 12.3 13.2 8.4 4.4 10.3 8.8 4.8 10.9 15.9 9.7 4.5 11.5 5.7 10.8 9.7 8.2 10.3 12.2 10.6 16.2 15.2 1.7 11.7 11.9 10.0 12.4
X[bar]= 10.41
S= 3.71
[10.41±1.645*]
[9.26; 11.56]
Using a confidence level of 90% you'd expect that the interval [9.26; 11.56]% contains the value of the population mean of the percentage of absences per tutorial of the students over the past 5 years.
I hope this helps!