What is 11.510510510 as a fraction

Replace ∗ with a monomial so that the result is an identity:(5x + ∗)(5x − ∗) = 25x2–0.16y4

Eddi Din [679]

Answer:

0.4y²

Step-by-step explanation:

25x² - 0.16y⁴

(5x)² - (0.4y²)²

(5x - 0.4y²)(5x + 0.4y²)

5 0

3 years ago

Maria kept a log of the number of miles she jogged each time she went for a run as she trained for an upcoming marathon. She dis

zhuklara [117]

Answer: The answer is 1 to 7.99.

7 0

4 years ago

Read 2 more answers

A new post-surgical treatment is being compared with a standard treatment. Seven subjects receive the new treatment, while seven

AnnyKZ [126]

Answer:

$t=\frac{17.71-28.71}{\sqrt{\frac{4.461^2}{7}+\frac{7.387^2}{7}}}}=-3.372$

$df=n_{S}+n_{N}-2=7+7-2=12$

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

$p_v =P(t_{(12)}$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean time for the new treatment is significantly lower than the time for the standard treatment at 5% of significance.

Step-by-step explanation:

New: 12, 13,15,19,20,21

Standard: 18,23,24,30,32,35

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

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

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

Data given and notation

$\bar X_{N}=17.71$ represent the mean for the new case

$\bar X_{S}=28.71$ represent the mean for the standard case

$s_{N}=4.461$ represent the sample standard deviation for the new case

$s_{S}=7.387$ represent the sample standard deviation for the standard case

$n_{S}=7$ sample size selected for the standard case

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

$\alpha=0.05$ 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 recovery time for patients who receive the new treatment is less than the mean for patients who receive the standard treatment, the system of hypothesis would be:

Null hypothesis: $\mu_{N} \geq \mu_{S}$

Alternative hypothesis: $\mu_{N} < \mu_{S}$

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:

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

$t=\frac{17.71-28.71}{\sqrt{\frac{4.461^2}{7}+\frac{7.387^2}{7}}}}=-3.372$

P-value

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

$df=n_{S}+n_{N}-2=7+7-2=12$

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

$p_v =P(t_{(12)}$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean time for the new treatment is significantly lower than the time for the standard treatment at 5% of significance.

6 0

4 years ago

Solve for n.<br> d=5m +11n

Serhud [2]

Answer:

$n=-\frac{5m}{11} +\frac{d}{11}$ or $n=\frac{d}{11}-\frac{5m}{11}$