Fifteen Players , 2 coaches , anf 21 parents attened this year's sports banquey. The cost of each meal was $12.00 plus 5% (0.05)
1 answer:
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Answer:
There is no sufficient evidence to support the claim that the bags are underfilled
Step-by-step explanation:
We are given;
n = 27 bags
Sample mean;X = 402 gram
Population mean;μ = 404 gram
Variance = 676
We know that, standard deviation(σ) = √variance
Thus;
σ = √676
σ = 26
The hypotheses are;
Null hypothesis; H0: μ = 404
Alternative hypothesis; HA: μ < 404
Let's find the z-score from the formula;
z = (X - μ)/(√σ/n)
z = (402 - 404)/√(26/27)
z = -2.04
From the z-distribution table attached, we have a p-value of 0.02068
This is less than the significance value of 0.01 and thus we will reject the null hypothesis and conclude that there is no sufficient evidence to support the claim that the bags are underfilled
Answer:
The answer is "Options A, B, and E represent mutually exclusive events".
Step-by-step explanation:
Two occurrences that can happen immediately called mutually incompatible. Let's now glance at our options and figure out where the statements are mutually incompatible events.
In Option A: You could see that landing on an unwanted portion and arriving on 2 are events that are locally incompatible, even as undesirable portion contains 3 and 4, and 2 were shaded.
In Option B: Arriving on a shaded part and falling on 3 are also mutually incompatible because there are 3 on a windows azure.
In Option C: A darkened portion and an increasing amount can land while 2 would be an even number as well as on the shaded portion. That number is very much the same.
In Option D: At the same time as 4 is greater than 3 and it is situated upon an undistressed section, landing and attracting a number larger than 3 can happen.
In Option E: Landing on a shaded part and landing on even a shaded part is an excluding event, since shaders may either be shaded or unlit.
