1-9 = 9 digits
10-99 = 180 digits
So if we continue the pattern to 99, there are 189 digits, and the last 5 digits would be 79899. Counting backwards: 189th = 9, 188th = 9, 187th = 8, 186th = 9, 185th = 7.
The 185th digit is 7.
Hello there, and thank you for posting your question here on brainly.
To convert a decimal to a percent, you have to make the decimal a whole number by getting rid of the decimal point, and then at a % sign.
0.125 ===> 125
125 ===> 125%
Hope this helped!! ☺♥
Which monomial is a perfect cube?<span>16x6</span><span>27x8</span><span>32x12</span><span>64x<span>6
Its D </span></span>
Answer:
864.36 boxes
Step-by-step explanation:
In the question above, we are given the following values,
Confidence interval 95%
Since we know the confidence interval, we can find the score.
Z score = 1.96
σ , Standards deviation = 15mm
Margin of error = 1 mm
The formula to use to solve the above question is given as:
No of boxes =[ (z score × standard deviation)/ margin of error]²
No of boxes = [(1.96 × 15)/1]²
= 864.36 boxes
Based on the options above, we can round it up to 97 boxes.
Answer:
Step-by-step explanation:
Hello!
Given the linear regression of Y: "Annual salary" as a function of X: "Mean score on teaching evaluation" of a population of university professors. It is desired to study whether student evaluations are related to salaries.
The population equation line is
E(Y)= β₀ + β₁X
Using the information of a n= 100 sample, the following data was calculated:
R²= 0.23
Coefficient Standard Error
Intercept 25675.5 11393
x 5321 2119
The estimated equation is
^Y= 25675.5 + 5321X
Now if the interest is to test if the teaching evaluation affects the proffesor's annual salary, the hypotheses are:
H₀: β = 0
H₁: β ≠ 0
There are two statistic you can use to make this test, a Student's t or an ANOVA F.
Since you have information about the estimation of β you can calculate the two tailed t test using the formula:
~
= 25.1109
The p-value is two-tailed, and is the probability of getting a value as extreme as the calculated
under the distribution 
p-value < 0.00001
I hope it helps!