Answers:
13 42 m; B; 0.57 m
Step-by-step explanation:
Data:
Pool A: r = 22 ft
Pool B: D = 13.6 m
Calculations:
1. Radius of Pool A
r = 22 ft × (0.305 m/1 ft) = 6.71 m
2. Diameter of Pool A
D =2r = 2 × 6.71 = 13.42 m
The diameter of Pool A is 13.42 m.
3. Compare pool diameters
The diameter of Pool B is 13.6 m.
So, the diameter of Pool <u>B</u> is greater.
4. Compare circumferences
The formula for the circumference of a circle is
C = 2πr or C = πD
Pool A: C = 2π × 6.71 = 42.16 m
Pool B: r = π × 13.6 = 42.73 m
Pool B - Pool A = 42.73 - 42.16 = <u>0.57 m
</u>
The circumference is greater by <u>0.57 m.</u>
<h2>
Hello!</h2>
The answer is: 33.33%
<h2>Why?</h2>
Since we have the average number of accidents that occurs in 1 month, and it's equal to 3, we can calculate the probability of 1 accident occurs by dividing it into the average number of accidents, using the following formula:
Where,
Favorable outcomes are the occurrence of the event, for this case, it's equal to 1.
Outcomes are the possible occurrence of the event, for this case, it's equal to 3.
So, by substituting we have:
So, the probability will be equal to 33.33%
Have a nice day!
Answer:
Step-by-step explanation:
-5x - 10y = -30
5x + 3y = 2
-7y = -28
y = 4
x + 8 = 6
x = -2
(-2, 4)
Answer:
Equation Form: x=−2,y=−2
Step-by-step explanation:
Eliminate the equal sides of each equation and combine.
3/2x+1=−x−4
Solve 3/2x+1=−x−4
for x. x=−2
Evaluate y when x=−2.
y=−2
The solution to the system is the complete set of ordered pairs that are valid solutions.
(−2,−2)
The result can be shown in multiple forms.
Point Form:
(−2,−2)
Equation Form:
x=−2,y=−2
Answer:
An interval that will likely include the proportion of students in the population of twelfth-graders who carry more than $15 is 960
Step-by-step explanation:
For example, Condition 1: n(.05)≤N
• The sample size (10) is less than 5% of the population (millions of musicians), so
the condition is met.
• Condition 2: np(1-p)≥10
• =
2
10
= .2
• 1 − = 10 .2 1 − .2 = 1.6 . This is less than 10 so this condition is not
met.
It would not be practical to construct the confidence interval.
