Answer:
(0.55, 0.75)
Step-by-step explanation:
The range can be estimated to be 6 standard deviations wide. Therefore, the standard deviation is:
σ = (0.72 - 0.42) / 6
σ = 0.05
The margin of error is ±2σ, so:
ME = ±0.10
Therefore, the interval estimate is:
(0.65 - 0.10, 0.65 + 0.10)
(0.55, 0.75)
Given that,
The camera sights the stadium at a 7 degree angle of depression.
The altitude of the blimp i slide 300 m.
To find,
The line of sight distance from the camera to the stadium.
Solution,
If we consider a right angled triangle. Let x is its hypotenuse i.e. the line of sight distance from the camera to the stadium. Using trigonometry :

So, the line of sight is at a distance of 2461.65 m from the camera to the stadium.
Answer:
10 cm.
Step-by-step explanation:
We'll begin by calculating the area of the small bubble. This can be obtained as follow:
Radius (r) = 5 cm
Area (A) =?
Since the bubble is circular in nature, we shall use the formula for area of circle to determine the area of the bubble. This is illustrated below:
A = πr²
A = π × 5²
A = 25π cm²
Next, we shall determine the total area of the small bubbles. This can be obtained as follow:
Area of 1 bubble = 25π cm²
Therefore,
Area of 4 bubbles = 4 × 25π cm²
Area of 4 bubbles = 100π cm²
Finally, we shall determine the radius of the large bubble. This can be obtained as follow:
Area of large bubble = total area of small bubbles = 100π cm²
Radius (r) =?
A = πr²
100π = πr²
100 = r²
Take the square root of both side
r = √100
r = 10 cm
Thus, the radius of the large bubble is 10 cm
The answer is 13. Or the third choice, just took the test.
We can use the FOIL method to solve.
(5x + 3)(7x - 7)
(5x * 7x) + (5x * -7) + (3 * 7x) + (3 * -7)
35x - 35x + 21x - 21
0 + 0
0
Best of Luck!