Since the homecoming committee bought only 500 footballs,they can sell no more than this. Represent this constraint with an ineq
1 answer:
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Answer:
8th term is 179.15904
Step-by-step explanation:
The sequence is geometric with a common ration = 6/5
r = 60/50 = 72/60 = 6/5
Use the formula
In your case a = 50, r = 6/5 and n = 8
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%.