Write each phrase as an algebraic expression. 16 seconds faster than Aya’s time.
1 answer:
Given:
The given phrase is 16 seconds faster than Aya’s time.
To find:
The algebraic expression for the given phrase.
Solution:
Let x be the Aya’s time.
16 seconds faster than Aya’s time. It means 16 seconds less than Aya’s time.
We know that a negative sign "-" is used for less than.
16 seconds less than Aya’s time =
16 seconds faster than Aya’s time =
Therefore, the required algebraic expression is .
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Answer:
Part 1:
The solution set of the system of equations is x = 1, y = 1
Part 2:
The solution set is x = 1, y = 1
Part 3:
The similarity of the system of equations in Parts 1 and 2 is that they have the same solution set
The difference of the system of equations in Parts 1 and 2 is that they are arranged differently
Step-by-step explanation:
Part 1:
The system of equation is given as follows;
2·x + y = 3...(1)
x = 2·y - 1...(2)
The above system of equations can be written in terms of the variable, y, as follows;
For equation (1), we have;
y = 3 - 2·x
For equation (2), we have;
y = (x + 1)/2
From the attached graph created with Microsoft Excel, we have;
The solution set (the point of intersection) of the system of equations is x = 1, y = 1
To accurately find the common solution, we have;
(x + 1)/2 = 3 - 2·x
x + 1 = 2·(3 - 2·x) = 6 - 4·x
x + 1 = 6 - 4·x
x + 4·x = 6 - 1
5·x = 5
x = 5/5 = 1
x = 1
Therefore, y = 3 - 2·x = 3 - 2× 1 = 1, at the common solution
Part 2:
y = -2x + 3
x - 2y = -1
∴ x = 2y -1
y = -2(2y -1) + 3
y = -4y + 2 + 3
5y = 5
y = 1
x = 2y - 1 = 2 - 1 = 1
x = 1
The solution set is x = 1, y = 1
Part 3
The system of equations are similar in terms of their solution
The system of equations are different in terms of their arrangement
Answer:
Step 1: The null and alternative hypothesis for the one way analysis of variance is,
H₀ : There is no difference in the means of each area.
H₁ : At least one mean is different.
The objective of this question is to test whether the income of the three areas are significantly different.
Step 2: The problem states to use 0.05 significance level.
Step 3: The test statistic follows the F-distribution.
Step 4: Decision rule is based on the critical value.
Numerator degrees of freedom is,
K - 1 = 3 - 1
= 2
Denominator degrees of freedom is 2.
n - k = 12 - 3
= 9
Using the F-distribution table for ∝=0.05 with numerator and denominator degrees freedom, the critical value is 4.26. The decision rule is to reject H₀ if the computed value of F exceeds 4.26.
Step 5: Calculations.
Using Excel follow the steps mentioned below.
1. Import or type the data into spreadsheet.
2. Data--------> Data Analysis----------> Anova: Single Factor, click OK.
3. Select Input Range, make a tick mark on Labels in the first row, enter 0.05 for alfa.
4. Click OK.
The obtained output for the one-way ANOVA is,
[ find the figure in attachment]
From the obtained output, the computed test statistic value, F = 14.18 , which is greater than the critical value of 4.26. So, reject H₀
Step 6: Interpretation. The average income of each area is different. It can be concluded that the average income is significantly different .
