George had a score of 25 points in a game. During the rest of the game, he lost 59 points. What score did George have at the end
1 answer:
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Answer:
2 solutions
Step-by-step explanation:
I like to use a graphing calculator to find solutions for equations like these. The two solutions are ...
- x = 1
- x = 4
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To solve this algebraically, it is convenient to subtract 2x-7 from both sides of the equation:
3x(x -4) +5 -x -(2x -7) = 0
3x^2 -12x +5 -x -2x +7 = 0 . . . . . eliminate parentheses
3x^2 -15x +12 = 0 . . . . . . . . . . . . collect terms
3(x -1)(x -4) = 0 . . . . . . . . . . . . . . . factor
The values of x that make these factors zero are x=1 and x=4. These are the solutions to the equation. There are two solutions.
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<em>Alternate method</em>
Once you get to the quadratic form, you can find the number of solutions without actually finding the solutions. The discriminant is ...
d = b^2 -4ac . . . . where a, b, c are the coefficients in the form ax^2+bx+c
d = (-15)^2 -4(3)(12) = 225 -144 = 81
This positive value means the equation has 2 real solutions.
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.