A pool has some initial amount of water in it. Then it starts being filled so the water level rises at a rate of 6 cm per minute
. After 20min the water level is 220 cm
1 answer:
Answer:
Step-by-step explanation:
I think the question below is the one you wanted to ask:
<em>A pool has some initial amount of water in it. Then it starts being filled so the water level rises at a rate of 6 centimeters per minute. After 20 minutes, the water level is 220 centimeters </em>
<em>Graph the pool's water level (in centimeters) as a function of time (in minutes).</em>
Here is my answer:
To draw the linear line, we need to identify the equation that modles this situation. As we know the water level rises at a rate of 6 cm per minute which mean the slope m = 6
The standard form is: y = mx + b <=> y = 6x + b
After 20min the water level is 220 cm
<=> 220 = 6(20) +b
<=> b = 100
So your equation is : y = 6x + 100 and the initial amount of water in it is 100
Please have a look at the attached photo
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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!