Answer:
1. -7.5
2. $1
3. 40
Step-by-step explanation:
For number 1, it can be solved by using the PEMDAS method, or see explanation below:
6x - 4x - 36 = 6 - 2x
2x - 36= -2x + 6
4x + 36 = 6
4x = -30
x = -15/2 or -7.5
For number 2, substitute 3 into both equations:
f(x) =1.50(3) + 2.00
and
f(x) = 2.00(3) + 1.50
This would get $6.50 and $7.50, which, if subtracted, gets $1.
For number 3, do something similar to the previous problem. Substitute 3 for x. It would be 5(2^3), or 40.
Hope this helps!
Answer:
Radius = 5 units
Step-by-step explanation:
Diameter of a circle = 10
Radius = diameter/2
Radius = 10/2
Radius = 5 units
A rational number is: A number that can be written as a quotient of 2 integers, where the divisor is not 0.
Example: 2/5 , -6,
Because you can write any integer as a quotient with a denominator of 1, all integers are rational numbers,
Example: 5 = 5/1
~Aamira~
Hope this helped☺☺
<h3>Answer:</h3>
length: 7 ft
width: 5 ft
<h3>Explanation:</h3>
Your familiarity with times tables tells you that 35 = 5×7. Checking, you find that 7 = 3×5 - 8, so you know that these values (5, 7) are the width and length of the rectangle (in feet).
Answer:
Null hypothesis: ![\mu_1 = \mu_2 = ..... \mu_j , j =1,2,....,n](https://tex.z-dn.net/?f=%20%5Cmu_1%20%3D%20%5Cmu_2%20%3D%20.....%20%5Cmu_j%20%2C%20j%20%3D1%2C2%2C....%2Cn)
Alternative hypothesis: ![\mu_i \neq \mu_j , i,j =1,2,....,n](https://tex.z-dn.net/?f=%20%5Cmu_i%20%5Cneq%20%5Cmu_j%20%2C%20i%2Cj%20%3D1%2C2%2C....%2Cn)
The alternative hypothesis for this case is that at least one mean is different from the others.
And the best method for this case is an ANOVA test.
Step-by-step explanation:
For this case we wnat to test if all the mean length of all face-to-face meetings and the mean length of all Zoom meetings are the same. So then the system of hypothesis are:
Null hypothesis: ![\mu_1 = \mu_2 = ..... \mu_j , j =1,2,....,n](https://tex.z-dn.net/?f=%20%5Cmu_1%20%3D%20%5Cmu_2%20%3D%20.....%20%5Cmu_j%20%2C%20j%20%3D1%2C2%2C....%2Cn)
Alternative hypothesis: ![\mu_i \neq \mu_j , i,j =1,2,....,n](https://tex.z-dn.net/?f=%20%5Cmu_i%20%5Cneq%20%5Cmu_j%20%2C%20i%2Cj%20%3D1%2C2%2C....%2Cn)
The alternative hypothesis for this case is that at least one mean is different from the others.
And the best method for this case is an ANOVA test.