Please help with that two math problems. I'll give brainly!
1 answer:
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Answer:
- Part (a) Vertex = (-3, -1)
- Part (b) Vertex = (4, 5)
- Part (c) function 2; max value = 5
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Explanations:
Part (a)
In this case, the vertex is the highest point, so the vertex of function 1 is (-3, -1)
If the parabola was flipped so that it opened upward, instead of downward, then the vertex would be the lowest point.
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Part (b)
We could graph this function to see where the highest or lowest point is, but there's another way we could find the vertex using algebra.
The given equation is in the form y = ax^2 + bx + c where in this case
- a = -2
- b = 16
- c = -27
The a and b values are plugged into the formula below to get
h = -b/(2a)
h = -16/(2*(-2))
h = -16/(-4)
h = 4
This is the x coordinate the vertex since the vertex is (h,k)
To find k, we plug that value of h into the original function to find its corresponding y value.
y = -2x^2 + 16x - 27
y = -2(4)^2 + 16(4) - 27
y = -2(16) + 16(4) - 27
y = -32 + 64 - 27
y = 5
We found that h = 4 leads to k = 5. The vertex is (h,k) = (4, 5)
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Part (c)
We'll compare the y values of each vertex
- Function 1 has a vertex of (-3, 1)
- Function 2 has a vertex of (4, 5)
The vertex with the larger y value is the winner, so that would be function 2. This parabola reaches a higher peak compared to the other curve.
The max value here is y = 5. In other words, it's the maximum output the function can produce.
If you graphed functions 1 and 2 together on the same xy grid (see below), you'll see that the curve for function 2 reaches a higher peak.
Answer:
Part A
b. 14.6 ± 7.38
Part B
b. 3.43
Part C
a. P-value < 0.01
Part D
b. There is sufficient evidence to reject the null hypothesis
Step-by-step explanation:
Part A
The given data are;
The number of seedlings in the field = 20
The number of seedlings selected to receive herbicide A = 10
The number of seedlings selected to receive herbicide B = 10
The height in centimeters of seedlings treated with Herbicide A, = 94.5 cm
The standard deviation, s₁ = 10 cm
The height in centimeters of seedlings treated with Herbicide B, = 109.1 cm
The standard deviation, s₂ = 9 cm
The 90% confidence interval for μ₂ - μ₁, is given as follows;
The critical-t at 95% and n₁ + n₂ - 2 degrees of freedom is given as follows;
The degrees of freedom, df = n₁ + n₂ - 2 = 10 + 10 - 2 = 18
α = 100% - 90% = 10%
∴ For two tailed test, we have, α/2 = 10%/2 = 5% = 0.05
= 1.734
C.I. ≈ 14.6 ± 7.37714603353
The 90% C.I. ≈ 14.6 ± 7.38
b. 14.6 ± 7.38
Part B
With the hypotheses are given as follows;
H₀; μ₂ - μ₁ = 0
Hₐ; μ₂ - μ₁ ≠ 0
The two sample t-statistic is given as follows;
The two sample t-statistic ≈ 3.43
b. 3.43
Part C
From the t-table, the p-value, we have, the p-value < 0.01
a. P-value < 0.01
Part D
Given that a significance level of 0.05 level is used and the p-value of 0.01 is less than the significance level, there is enough statistical evidence to reject the null hypothesis
b. There is sufficient evidence to reject the null hypothesis.