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This year, gasoline costs $4 per gallon. This is a 125% increase over the cost 3 years ago. What expression can be used to deter

pochemuha

The equation would be, 4/125%.

7 0

4 years ago

The sum of a number and six is seven times the number

olga_2 [115]

The answer is D. I hope this helps love! :)

7 0

3 years ago

Quantitative noninvasive techniques are needed for routinely assessing symptoms of peripheral neuropathies, such as carpal tunne

Pavlova-9 [17]

Answer:

$t=\frac{2.35-1.83}{\sqrt{\frac{0.88^2}{10}+\frac{0.54^2}{7}}}}=1.507$

$p_v =P(t_{(15)}>1.507)=0.076$

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the group CTS, NOT have a significant higher mean compared to the Normal group at 1% of significance.

Step-by-step explanation:

1) Data given and notation

$\bar X_{CTS}=2.35$ represent the mean for the sample CTS

$\bar X_{N}=1.83$ represent the mean for the sample Normal

$s_{CTS}=0.88$ represent the sample standard deviation for the sample of CTS

$s_{N}=0.54$ represent the sample standard deviation for the sample of Normal

$n_{CTS}=10$ sample size selected for the CTS

$n_{N}=7$ sample size selected for the Normal

$\alpha=0.01$ 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)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean for the group CTS is higher than the mean for the Normal, the system of hypothesis would be:

Null hypothesis: $\mu_{CTS} \leq \mu_{N}$

Alternative hypothesis: $\mu_{CTS} > \mu_{N}$

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:

$t=\frac{\bar X_{CTS}-\bar X_{N}}{\sqrt{\frac{s^2_{CTS}}{n_{CTS}}+\frac{s^2_{N}}{n_{N}}}}$ (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:

$t=\frac{2.35-1.83}{\sqrt{\frac{0.88^2}{10}+\frac{0.54^2}{7}}}}=1.507$

P-value

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

$df=n_{CTS}+n_{N}-2=10+7-2=15$

Since is a one side right tailed test the p value would be:

$p_v =P(t_{(15)}>1.507)=0.076$

We can use the following excel code to calculate the p value in Excel:"=1-T.DIST(1.507,15,TRUE)"

Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the group CTS, NOT have a significant higher mean compared to the Normal group at 1% of significance.

6 0

4 years ago

The graph of y=3x^2-3x-1 is shown. Use the graph to find estimates for the solutions of i)3x^2-3x+2=2 ii) 3x^2-3x-1=x+1

AlexFokin [52]

Answer:

i) (0, 2) and (1, 2), ii) (0.333, 1.333) and (1, 2).

Step-by-step explanation:

i) Let be $y=3\cdot x^{2}-3\cdot x -1$ , if $3\cdot x^{2}-3\cdot x +2 = 2$ , which is equivalent to the following system of equations:

$y = 3\cdot x^{2}-3\cdot x +2$

$y = 2$

Now, this system is now represented by means of a graphing tool and whose outcome is attached below. There are two solutions: (0, 2) and (1, 2)

ii) Let be $y=3\cdot x^{2}-3\cdot x -1$ , if $3\cdot x^{2}-3\cdot x +2 = x+1$ , which is equivalent to the following system of equations:

$y = 3\cdot x^{2}-3\cdot x +2$

$y = x+1$

Now, this system is now represented by means of a graphing tool and whose outcome is attached below. There are two solutions: (0.333, 1.333) and (1, 2)

7 0

4 years ago

A $60 shirt is on sale for 30% off. How much is the shirt's sale price?

vlada-n [284]

Answer:39?

Step-by-step explanation:

4 0

3 years ago

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