Answer:
c= 25+0.05m
Step-by-step explanation:
Given that,
The phone company charges a flat rate of $25 per month. In addition they charge $0.05 for each minute of service.
$25 is fixed here and charge $0.05 for each minute of service.
We need to find the equation that can be used to find the monthly charge based upon the number of minutes (m) of service each month.
c= 25+0.05m
Hence, this is the required equation.
35 squared + 35 squared
rad 2450
35 rad 2
Answer:
1. Her weekly gross is $234
2. $421.20 for federal income tax withholding.
3. $174.096 for social security tax
4. $40.716 for medical tax
5. Her net paycheck is $636.012
6. $2808 for her total gross pay
7. $2171.988 is her total net pay
The expected value of the amount of average snowfall for over 30 years is 86.7 inches with a standard deviation of 40.4 inches. To verify if this particular trend continues, we must check the significance value of the amount snowfall for the past four years.
Given that the snowfall for past years are as follows: 115.7 inches, 62.9 inches, 168.5 inches, and 135.7 inches.
Thus the mean of the sample would be: (115.7 + 62.9 + 168.5 + 135.7)/4 = 120.7 inches.
To compute for the z-score, we have
z-score = (x – μ) / (σ / √n)
where x is the computed/measured value, μ is the expected mean, σ is the standard deviation, and n is the number of samples.
Using the information we have,
z-score (z) = (120.7 - 86.7) / (40.4/ √4) = 1.68
In order to reject the null hyptohesis our probability value must be less than the significance level of 5%. For our case, since z = 1.68, P-value = 0.093 > 0.05.
Therefore, the answer is B.
Answer:
<em>A) (-5,7)</em>
Step-by-step explanation:
<u>Functions and Relations</u>
A set of values A can have a relation with another set B as long as at least one element of A has at least one image in B. Functions are special relations where each element of A (the domain of the function) has one and only one image on B (the range of the function).
By looking at the options, we can see that x=9, x=-8, and x=-1 already have defined values in Y, so if we define another value for any of them the relation will stop being a function. The only possible choice to preserve the function is the option
