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slava [35]
3 years ago
14

A satellite camera takes a rectangular-shaped picture. The smallest region that can be photographed is a 4-km by 4-km rectangle.

As the camera zooms out, the length l and width w of the rectangle increase at a rate of 3 km/sec. How long does it take for the area A to be at least 4 times its original size?
Mathematics
1 answer:
Lesechka [4]3 years ago
6 0
So we know that the formula for the area of a rectangle is A = lw.
Now both the length and width of the rectangle increase at 3 km/s, therefore, A(t) = (3t+l)*(3t+w). Since the initial length = initial width = 4 km, then the initial area = 16 [tex]km^2. We want to know the time when the area is four times its original area, therefore, our new formula is: 4A(t) = (3t+l)*(3t+w). Plugging in our known values we have:

64 [km^2] = (3t + 4 [km])*(3t + 4 [km])
t =  \frac{4}{3} s

The area is four times its original area after <span>\frac{4}{3} s[/tex]</span>.
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The 99% confidence interval of the population standard deviation is 1.7047 < σ < 7.485

Step-by-step explanation:

Confidence interval of standard deviation is given as follows;

\sqrt{\dfrac{\left (n-1  \right )s^{2}}{\chi _{1-\alpha /2}^{}}}< \sigma < \sqrt{\dfrac{\left (n-1  \right )s^{2}}{\chi _{\alpha /2}^{}}}

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χ = Chi squared value at the given confidence level

\bar x = ∑x/n = (62 + 58 + 58 + 56 + 60 +53 + 58)/7 = 57.857

The sample standard deviation s = \sqrt{\dfrac{\Sigma (x - \bar x)^2}{n - 1} } = 2.854

The test statistic, derived through computation, = ±3.707

Which gives;

C. I. = 57.857 \pm 3.707 \times \dfrac{2.854}{\sqrt{7} }

\sqrt{\dfrac{\left (7-1  \right )2.854^{2}}{16.812}^{}}}< \sigma < \sqrt{\dfrac{\left (7-1  \right )2.854^{2}}{0.872}}

1.7047 < σ < 7.485

The 99% confidence interval of the population standard deviation = 1.7047 < σ < 7.485.

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