Approximate 5.7255 to the nearest thousand

A line with a slope of -4 passes through the point (0, 0). What is its equation in

Dahasolnce [82]

Answer: y = -4x

Step-by-step explanation: You would use the formula y = mx + b to solve your problem. When you plug the numbers in, you get y = -4x. There would be no "+ b" because in this case, the y-intercept (b) is equal to zero. Therefore, you would not need to write "y = -4x + 0" because that would not be the equation in its simplest form.

4 0

2 years ago

Find vertex and y- intercept and x-intercept in x^2 -2x-15

AfilCa [17]

It would be (x-5)(x+3)

4 0

2 years ago

The distribution of the amount of money spent by students on textbooks in a semester is approximately normal in shape with a mea

zheka24 [161]

Answer:

$301 - $397

Step-by-step explanation:

Using the Empirical rule

1) 68% of data falls within 1 standard deviation from the mean - that means between μ - σ and μ + σ .

2)95% of data falls within 2 standard deviations from the mean - between μ – 2σ and μ + 2σ .

3)99.7% of data falls within 3 standard deviations from the mean - between μ - 3σ and μ + 3σ .

From the above question,

Mean = 349 , standard deviation = 24.

Confidence interval = 95%

Using 2)95% of data falls within 2 standard deviations from the mean - between μ – 2σ and μ + 2σ .

μ – 2σ

= 349 - 2(24)

= 349 - 48

= 301

μ + 2σ

349 + 2(24)

= 349 + 48

= 397

Therefore, according to the standard deviation rule, approximately 95% of the students spent between $301 and $397 on textbooks in a semester.

8 0

2 years ago

(17,11) (5,0) find the slope show the work

Allushta [10]

0-11/5-17=11/12 is the answer to your question. The work is the equation.

4 0

2 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

3 years ago