A. (2 + 4) + 5( − 1) − ( + 7)
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Answer:
The probability of the stick's weight being 2.33 oz or greater is 0.0041 or 0.41%.
Step-by-step explanation:
Given:
Weight of a given sample (x) = 2.33 oz
Mean weight (μ) = 1.75 oz
Standard deviation (σ) = 0.22 oz
The distribution is normal distribution.
So, first, we will find the z-score of the distribution using the formula:
Plug in the values and solve for 'z'. This gives,
So, the z-score of the distribution is 2.64.
Now, we need the probability .
From the normal distribution table for z-score equal to 2.64, the value of the probability is 0.9959. This is the area to the left of the curve or less than z-score value.
But, we need area more than the z-score value. So, the area is:
Therefore, the probability of the stick's weight being 2.33 oz or greater is 0.0041 or 0.41%.
Reflection across the line y=x simply swaps the x and y coordinates.
K'(-2, -5), A'(1, -4), I'(-1, 0), J'(-4, -2)
K'(-2, -5), A'(1, -4), I'(-1, 0), J'(-4, -2)