Answer:
Here is the graph:
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Factorise 5x + 40 7x - 49 18x + 9 15x - 10
1 answer:
<em><u>Factorization is:</u></em>
Given that we have to factor the given expressions
<h3><u>First expression:</u></h3>5x + 40
We have to take the greatest common factor out
When we find all the factors of two or more numbers, and some factors are the same then the largest of those common factors is the Greatest Common Factor
Take G.C.F of 5 and 40
The factors of 5 are: 1, 5
The factors of 40 are: 1, 2, 4, 5, 8, 10, 20, 40
Then the greatest common factor is 5
Factor out 5 from given expression
7x - 49
Similarly, take G.C.F of 7 and 49
The factors of 7 are: 1, 7
The factors of 49 are: 1, 7, 49
Then the greatest common factor is 7
Factor out 7 from given expression
18x + 9
Similarly, take G.C.F of 18 and 9
The factors of 9 are: 1, 3, 9
The factors of 18 are: 1, 2, 3, 6, 9, 18
Then the greatest common factor is 9
Factor out 9 from given expression
15x - 10
Similarly, take G.C.F of 15 and 10
The factors of 10 are: 1, 2, 5, 10
The factors of 15 are: 1, 3, 5, 15
Then the greatest common factor is 5
Factor out 5 from given expression
Thus factorization is found
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Answer:
1. M=108
2. μ=110
3. In the explanation.
4. Test statistic t = -1.05
5. P-value = 0.1597
Step-by-step explanation:
The question is incomplete: to solve this problem, we need the sample information: size, mean and standard deviation.
We will assume a sample size of 10 matches, a sample mean of 108 points and a sample standard deviation of 6 points.
1. The mean points is the sample points and has a value of 108 points.
2. The null hypothesis is H0: μ=110, meaning that the mean score is not significantly less from 110 points.
3. This is a hypothesis test for the population mean.
The claim is that the mean score is significantly less than 110.
Then, the null and alternative hypothesis are:
The significance level is 0.05.
The sample has a size n=10.
The sample mean is M=108.
As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=6.
The estimated standard error of the mean is computed using the formula:
4. Then, we can calculate the t-statistic as:
The degrees of freedom for this sample size are:
5. This test is a left-tailed test, with 9 degrees of freedom and t=-1.05, so the P-value for this test is calculated as (using a t-table):
As the P-value (0.1597) is bigger than the significance level (0.05), the effect is not significant.
The null hypothesis failed to be rejected.
There is not enough evidence to support the claim that the mean score is significantly less than 110.