Answer:
!9 degrees 6 point 4
Step-by-step explanation:
Answer:
(3,0,0)
Step-by-step explanation:
3x + 5y + z = 9 --------------(I)
-4y - z = 0 -------------------(II)
z = -4y
2x + 3y - 4z = 6 (III)
Substitute z = -4y in equation (I) and (III)
3x +5y - 4y = 9
3x + y = 9 -----------------------(a)
2x + 3y - 4*(-4y) = 6
2x + 3y + 16y = 6
2x + 19y = 6 -------------------(b)
Multiply equation (a) by (-19) and then add. So, y will be eliminated and we can get the value of x
(a)*(-19) -57x - 19y = -171
(b) <u> 2x + 19y = 6 </u>
-55x = -165
x = -165/-55
x = 3
Plugin x = 3 in equation (a)
3*3 + y = 9
9 + y = 9
y = 9 - 9
y = 0
Plug in y = 0 in equation (II)
-4*0 - z = 0
z = 0
Answer:
5x - y = - 2
Step-by-step explanation:
The equation of a line in standard form is
Ax + By = C ( A is a positive integer and B, C are integers )
First obtain the equation in slope- intercept form
y = mx + c ( m is the slope and c the y- intercept )
here m = 5 and the line passes through (0, 2) ⇒ c = 2
y = 5x + 2 ← in slope- intercept form
Subtract y from both sides
0 = 5x - y + 2 ( subtract 2 from both sides )
5x - y = - 2 ← in standard form
Answer:
You got it right.
Step-by-step explanation:
Answer:
Null hypothesis:
Alternative hypothesis:
The statistic would be given by:
(1)
Now we need to find the critical value for the rejection zone of the null hypothesis. Since we have a left tailed test we need to find in the normal standard distirbution a value who accumulate 0.05 of the area in the left tail and we got:
And the best choice for this case would be:
z < − 1.65
Step-by-step explanation:
Information provided
n=2431 represent the random sample taken
estimated proportion of interest
is the value that we want to test
represent the significance level
Confidence=95% or 0.95
z would represent the statistic
Hypothesis to test
We want to verify if the true proportion of Americans who reduced meat consumption last year is less than 0.25, then the system of hypothesis are :
Null hypothesis:
Alternative hypothesis:
The statistic would be given by:
(1)
Now we need to find the critical value for the rejection zone of the null hypothesis. Since we have a left tailed test we need to find in the normal standard distirbution a value who accumulate 0.05 of the area in the left tail and we got:
And the best choice for this case would be:
z < − 1.65