-2/3X + 4/3 = -10/3 solve each equation by first eliminating all fractions
1 answer:
The solution of -2/3X + 4/3 = -10/3 by first eliminating all fractions is 7.0006
<u>Solution:</u>
We have been given an equation as follows:
And we have been asked to solve each equation by first eliminating all fractions.
The only way to solve this equation is by converting each of its components into decimals since it is stated in the question that we should not solve it in fraction form.
To do so we will individually convert the fractions to decimals and then substitute them back into the equation.
The 1st fraction in decimal form is:
The 2nd fraction in decimal form is:
And the 3rd fraction in decimal form is:
So, now we substitute the decimals instead of the fractions in the given equation as follows:
Therefore, the value of x in decimal from is 7.0006
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Answer:
a) Null hypothesis:
Alternative hypothesis:
b) the test statistics is : 2.15
c) The p-value is 0.0158
d) NO, there is evidence that the mean time of telephone survey is less than 15 and premium rate is not justified.
Step-by-step explanation:
The data in the Microsoft Excel are:
17;11;12;23;20;23;15;
16;23;22;18;23;25;14;
12;12;20;18;12;19;11;
11;20;21;11;18;14;13;
13;19; 16;10;22;18;23.
a) Formulate the null and alternative hypotheses for this application.
From the question, Fowle Marketing Research Inc. is taking base charge from a client on the given assumption that if the mean time of telephone survey is 15 minutes or less.
The null and alternative hypotheses are therefore as follows:
Null hypothesis:
The null hypothesis states that there is evidence that the mean time of telephone survey is less than 15 and premium rate is not justified.
Alternative hypothesis:
The alternative hypothesis states that there is evidence that the mean time of telephone survey exceeds 15 and premium rate is justified.
b) Compute the value of the test statistic.
Given that:
n = 35
The sample mean is;
Thus:
Thus; the test statistics is : 2.15
c) What is the p-value?
p-value = P(Z > 2.15)
p-value = 1 - P(Z ≤ 2.15)
From the standard normal table, the value of P(Z ≤ 2.15) is 0.9842
p-value = 1 - 0.9842
p-value = 0.0158
The p-value is 0.0158
d) At a = .01, what is your conclusion?
According to the rejection rule, if p-value is less than 0.01 then reject null hypothesis at ∝ = 0.01
Thus; p-value = 0.0158 > ∝ = 0.01
By the rejection rule, accept the null hypothesis.
Therefore, there is evidence that the mean time of telephone survey is less than 15 and premium rate is not justified.