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kozerog [31]
3 years ago
12

Let h(x) = -4x + 2. h(5) =

Mathematics
2 answers:
Novay_Z [31]3 years ago
6 0

Answer:

h(5) =-18

Step-by-step explanation:

h(x) = -4x + 2

Let x=5

h(5) = -4*5 +2

      = -20+2

      = -18

borishaifa [10]3 years ago
4 0

Answer:

h(5)=-18

Step-by-step explanation:

h(5) means you'd be substituting the x for 5, so -4(5) + 2 is the same as -20+2 which is -18.

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t=\frac{19.1-23.3}{\sqrt{\frac{4.818^2}{10}+\frac{5.559^2}{10}}}}=-1.805  

p_v =P(t_{(18)}

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Step-by-step explanation:

Data given and notation

We can calculate the sample mean and deviation with these formulas:

\bar X = \frac{\sum_{i=1}^n X_i}{n}

s= \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

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n_{1}=10 sample size selected 1

n_{2}=10 sample size selected 2

\alpha represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

p_v represent the p value for the test (variable of interest)

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We need to conduct a hypothesis in order to check if the average time taken when training under method 1 is less than the average time for Method 2, the system of hypothesis would be:

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Alternative hypothesis:\mu_{1} < \mu_{2}

If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:

t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}} (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

We can replace in formula (1) the info given like this:

t=\frac{19.1-23.3}{\sqrt{\frac{4.818^2}{10}+\frac{5.559^2}{10}}}}=-1.805  

P-value

The first step is calculate the degrees of freedom, on this case:

df=n_{1}+n_{2}-2=10+10-2=18

Since is a one sided test the p value would be:

p_v =P(t_{(18)}

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If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude the the true mean for method 1 is lower than the mean for the method 2 at 5% of significance

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