Answer:
x = 20
Step-by-step explanation:
Traingle ABC = Traingle DEC
4x-1×4×5 = x+2×4×5
80x-1 = 2×20x
80x-20x = 2+1
60x = 3
60÷3 = x
x = 20
Answer:
The 95% confidence interval for the percentage of all boards in this shipment that fall outside the specification is (1.8%, 6.2%).
Step-by-step explanation:
In a random sample of 300 boards the number of boards that fall outside the specification is 12.
Compute the sample proportion of boards that fall outside the specification in this sample as follows:
The (1 - <em>α</em>)% confidence interval for population proportion <em>p</em> is:
The critical value of <em>z</em> for 95% confidence level is,
*Use a <em>z</em>-table.
Compute the 95% confidence interval for the proportion of all boards in this shipment that fall outside the specification as follows:
Thus, the 95% confidence interval for the proportion of all boards in this shipment that fall outside the specification is (1.8%, 6.2%).
Answer:
1) No
2) Yes
3)No
4)Yes
Step-by-step explanation:
1) 3(8) is simplified to 24
But 24 < 24 is not a true statement because they are both equal.
2) 3(5) is simplified to 15
15 < 40 is a true statement because 15 is less than 24
3) 3(11) simplifies to 33
33 < 24 is not a true statement because 33 is more than 24
4) 3(6) simplifies to 18
18 < 24 is a true statement because 18 is less than 24
Hope this helped.
Answer:
16x^4+96x^3y+216x^2 y^2+216xy^3+81y^4
Answer:
Step-by-step explanation:
Hello!
Given the variables
Y: standardized history test score in third grade.
X₁: final percentage in history class.
X₂: number of absences per student.
<em>Determine the following multiple regression values.</em>
I've estimated the multiple regression equation using statistics software:
^Y= a + b₁X₁ + b₂X₂
a= 118.68
b₁= 3.61
b₂= -3.61
^Y= 118.68 + 3.61X₁ - 3.61X₂
ANOVA Regression model:
Sum of Square:
SS regression: 25653.86
SS Total: 36819.23
F-ratio: 11.49
p-value: 0.0026
Se²= MMError= 1116.54
Hypothesis for the number of absences:
H₀: β₂=0
H₁: β₂≠0
Assuming α:0.05
p-value: 0.4645
The p-value is greater than the significance level, the decision is to not reject the null hypothesis. Then at 5% significance level, there is no evidence to reject the null hypothesis. You can conclude that there is no modification of the test score every time the number of absences increases one unit.
I hope this helps!
