Help me with this question please!
2 answers:
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Answer:
1) (2.2, -1.4)
2) (1.33, 1)
Step-by-step explanation:
Question 1)
Two lines, with their corresponding equations are given and we have to find the solution to the system of equations.
The given lines are:
Equation of Line 1:
This line passes through the points: (0, 2.5) , (2.2, -1.4)
Equation of Line 2:
This line passes through the points (0, -3) , (2.2, -1.4)
By looking at the graph/given data we have to find the solution of these linear equations.
Remember that the solution of linear equations is an ordered pair, through which both the lines pass i.e. the point at which both the given lines intersect is the solution of the linear equations.
From the given data we can see that both the lines pass through one common point, (2.2, -1.4). Since, both lines pass through this point, this means this is the point of intersection of the lines and hence there solution.
So, the answer to this questions is (2.2, -1.4)
Question 2)
The given equations are:
y = 1.5x - 1 Equation 1
y = 1 Equation 2
We can solve these equations by method of substitution.
Substituting the value of y from Equation 2, in Equation 1, we get:
1 = 1.5x - 1
1 + 1 = 1.5x
2 = 1.5x
x = 2/1.5
x = 1.33
y = 1
Thus, the solution of the given linear equations is (1.33, 1)
M∠X = 54.3°.
Using the Law of Sines, we have:
Cross multiplying gives us
61(sin 34) = 42(sin X)
Divide both sides by 42:
(61(sin 34))/42 = (42(sin X))/42
(61(sin 34))/42 = sin X
Take the inverse sine of both sides:
sin⁻¹((61(sin 34))/42) = sin⁻¹(sin X)
54.3 = X
Using the Law of Sines, we have:
Cross multiplying gives us
61(sin 34) = 42(sin X)
Divide both sides by 42:
(61(sin 34))/42 = (42(sin X))/42
(61(sin 34))/42 = sin X
Take the inverse sine of both sides:
sin⁻¹((61(sin 34))/42) = sin⁻¹(sin X)
54.3 = X
Answer:
Step-by-step explanation:
Hello!
You have two random samples obtained from two different normal populations.
Sample 1
n₁= 15
X[bar]₁= 350
S₁= 12
Sample 2
n₂= 17
X[bar]₂= 342
S₂= 15
At α: 0.05 you need to obtain the p-value for testing variances for a one tailed test.
If the statistic hypotheses are:
H₀: σ₁² ≥ σ₂²
H₁: σ₁² < σ₂²
The statistic to test the variances ratio is the Stenecor's-F test. ~
The p-value is:
P(≤0.64)= 0.02
I hope it helps!