Acceleration = vf-vi /t
10-22/3=2.6m/s^2
Answer:
97.5%
Explanation:
By the empirical rule (68-95-99.7),
- 68% of data are within <em>μ </em>- <em>σ</em> and <em>μ </em>+ <em>σ</em>
- 95% of data are within <em>μ </em>- 2<em>σ</em> and <em>μ </em>+ 2<em>σ</em>
- 99.7% of data are within <em>μ </em>- 3<em>σ</em> and <em>μ </em>+ 2<em>σ</em>
<em>σ </em> and <em>μ</em> are the standard deviation and the mean respectively.
From the question,
<em>μ</em> = 7.2 cm
<em>σ</em> = 0.38 cm
7.96 = 7.2 + (<em>n</em> × 0.38)
<em>n</em> = 2
Hence, 7.96 represents <em>μ </em>+ 2<em>σ</em>.
P(X < <em>μ </em>+ 2<em>σ</em>) = P(X < <em>μ</em>) + P(<em>μ</em> < X < <em>μ </em>+ 2<em>σ</em>)
P(X < <em>μ</em>) is the percentage less than the mean = 50%.
P(<em>μ</em> < X < <em>μ </em>+ 2<em>σ</em>) is half of P(<em>μ </em>- 2<em>σ</em> < X < <em>μ </em>+ 2<em>σ</em>) = 95% ÷ 2 = 47.5%.
Considering this, for apples that are no more than 7.96 cm,
P(X < 7.96) = P(X < 7.2) + P(7.2 < X < 7.96) = 50% + 47.5% = 97.5%
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Aperture is measured in F-stops, in which the f-stops is the amount of light allowed to pass through the aperture, which simply put means that the smaller the aperture, the higher the f-stops. What it does is reduce the amount of light that reaches the film, so the higher the f-stops, the less light reaches the film.
Health and safety are two reasons
