Solve equation -143 = 1+8(3 – 3b)
2 answers:
You might be interested in
Answer:
A. 4.5
Step-by-step explanation:
According to my graphing calculator, e^1.5=4.5 which is rounded to the nearest tenth.
Answer: e^1.5 = 4.482
Step-by-step explanation:
Given:
We need to find the value of function at x=1.5
Put x=1.5 into f(x)
Now we will verify the result using graph.
First we draw the graph of f(x) and a vertical line x=1.5
Please see the attachment for graph.
Hence, The value of
u - v = <5, 2> - <2, - 3> = <5 - 2, 2 + 3> = <3, 5>
given a vector <u, v> then the magnitude is √ (u² + v² ) and it's angle of direction is
θ = (
)
thus the magnitude of <3, 5> = √(3² + 5²) = √34 ≈ 5.83 ( to 2 dec. places )
Θ = (
) = 59.04° ( to 2 dec. places )
Answer:
a) There is a 10.75% probability of observing less than 60 hits in an hour.
b) The 99th percentile of the distribution of the number of hits is 95.21 hits.
c) There is a 24% probability of observing between 80 and 90 hits an hour
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
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. The sum of the probabilities is decimal 1. So 1-pvalue is the probability that the value of the measure is larger than X.
In this problem, we have that
The web logs of a certain website show that the average number of hits in an hour is 75 with a standard deviation equal to 8.6, so .
a) What’s the probability of observing less than 60 hits in an hour? Use the normal approximation
This is the pvalue of Z when X = 60. So
has a pvalue of 0.1075. This means that there is a 10.75% probability of observing less than 60 hits in an hour.
b) What’s the 99th percentile of the distribution of the number of hits?
What is the value of X when Z has a pvalue of 0.99.
Z = 2.35 has a pvalue of 0.99
So
The 99th percentile of the distribution of the number of hits is 95.21 hits.
c) What’s the probability of observing between 80 and 90 hits an hour?
This is the pvalue of the zscore of X = 90 subtracted by the pvalue of the zscore of X = 80.
For X = 90
has a pvalue of 0.95907
For X = 80
has a pvalue of 0.71904
So
There is a 0.95907 - 0.71904 = 0.24003 = 24% probability of observing between 80 and 90 hits an hour