Answer:
<h2>It is 159.1 in^2 </h2>
Step-by-step explanation:
I hope that is useful for you
Answer:
7.64% probability that they spend less than $160 on back-to-college electronics
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:

Probability that they spend less than $160 on back-to-college electronics
This is the pvalue of Z when X = 160. So



has a pvalue of 0.0763
7.64% probability that they spend less than $160 on back-to-college electronics
14t^3*6t
= (14*6)*(t^3*t)
= (2*7*2*3)*(t*t*t*t)
= 2*2*3*7*t*t*t*t
Answer:
yo slide me that number ong
Step-by-step explanation:
Explanation:
For the purpose of filling in the table, the BINOMPDF function is more appropriate. The table is asking for p(x)--not p(n≤x), which is what the CDF function gives you.
If you want to use the binomcdf function, the lower and upper limits should probably be the same: 0,0 or 1,1 or 2,2 and so on up to 5,5.
The binomcdf function on my TI-84 calculator only has the upper limit, so I would need to subtract the previous value to find the table entry for p(x).