4 formulas which include pi. 2 deal with circles and 2 deal with spheres
1 answer:
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Answer:
p-value: 0 .1292
Step-by-step explanation:
Hello!
The objective is to test if it is profitable to expand supply delivery. The company thinks that if more than 59% (symbolically p > 0.59) of the items are selling out in the markets, then it is profitable to increase the deliveries.
A sample of 48 markets was taken and it was registered that the item was sold out in 32 of them.
The study variable is.
X: Number of markets where the item was sold out in a random sample of 48 markets.
The study parameter is the proportion of "bare shelves"
sample proportion 'p= (32/48) = 0.67
The hypothesis is:
H₀: p ≤ 0.59
H₁: p > 0.59
α: 0.05
Remember: The p-value is defined as the probability corresponding to the calculated statistic if possible under the null hypothesis (i.e. the probability of obtaining a value as extreme as the value of the statistic under the null hypothesis).
So, to calculate the p-value you have to first calculate the statistic under the null hypothesis:
Z= 1.1269≅ 1.13
Keep in mind that the p-value as the test is one-tailed. Now you can calculate the p-value as:
P(Z ≥ 1.13)= 1 - P(Z < 1.13)= 1 - 0.8706 =0.1292
The decision is to reject the null hypothesis. So at a level of 5% you can say that it is probitable to increase the deliveries.
I hope you have a SUPER day!
Answer:
A = 0.75 gram or 1 gram
Step-by-step explanation:
The half-life of carbon 14 is years. How much would be left of an original -gram sample after 2,292 years? (To the nearest whole number).
We can use the following formula for half-life of to find out how much is left from the original sample after 2,292 years:
where:
<em>A</em> is the amount left of an original gram sample after <em>t</em> years, and
is the amount present at time <em>t</em> = 0.
The half-life of is the time <em>t</em> at which the amount present is one-half the amount at time <em>t </em>= 0.
If 1 gram of is present in a sample,
Solve for A when t = 2,292:
Substituting = 1 gram into the decay equation, and we have:
A = 0.75 g or 1 g