Input- -3 output= 2 what is three possible rule for this table?
1 answer:
You might be interested in
Answer:
a) 0.0392
b) 0.4688
c) At least $39,070 to be among the 5% most expensive.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
a. What is the probability that a wedding costs less than $20,000 (to 4 decimals)?
This is the pvalue of Z when X = 20000. So
has a pvalue of 0.0392.
So this probability is 0.0392.
b. What is the probability that a wedding costs between $20,000 and $30,000 (to 4 decimals)?
This is the pvalue of Z when X = 30000 subtracted by the pvalue of Z when X = 20000.
X = 30000
has a pvalue of 0.5080.
X = 20000
has a pvalue of 0.0392.
So this probability is 0.5080 - 0.0392 = 0.4688
c. For a wedding to be among the 5% most expensive, how much would it have to cost (to the nearest whole number)?
This is the value of X when Z has a pvalue of 0.95. So this is X when Z = 1.645.
The wedding would have to cost at least $39,070 to be among the 5% most expensive.
Answer:
c+s < 12, 2c+4s >= 30.
Step-by-step explanation:
Known: at least 30 legs, less than 12 heads.
Each animal has one head, so if there are less than 12 heads, then c + s < 12.
Sheep have four legs and chickens have two legs, so if there are at least 30 legs, then # of chicken legs + # of sheep legs >= 30, ----> c(2) + s(4) >= 30, ----> 2c + 4s >= 30.
The graph attached models the problem is you set the x axis as "c" (# of chickens) and y axis as "s" (# of sheep).
