Given f(x) = 3 and g(x) = cos(x). What is Limit of left-bracket f (x) minus g (x) right-bracket as x approaches negative pi?

A researcher was interested in comparing the resting pulse rates of people who exercise regularly and people who do not exercise

DiKsa [7]

Answer:

We Reject H₀ if t calculated > t tabulated

But in this case,

0.83 is not greater than 2.056

Therefore, we failed to reject H₀

There is no difference between the mean pulse rate of people who do not exercise and the mean pulse rate of people who do exercise.

Step-by-step explanation:

Refer to the attached data.

The Null and Alternate hypothesis is given by

Null hypotheses = H₀: μ₁ = μ₂

Alternate hypotheses = H₁: μ₁ ≠ μ₂

The test statistic is given by

$t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} } }$

Where $\bar{x}_1$ is the sample mean of people who do not exercise regularly.

Where $\bar{x}_2$ is the sample mean of people who do exercise regularly.

Where $s_1$ is the sample standard deviation of people who do not exercise regularly.

Where $s_2$ is the sample standard deviation of people who do exercise regularly.

Where $n_1$ is the sample size of people who do not exercise regularly.

Where $n_2$ is the sample size of people who do exercise regularly.

$t = \frac{72.7 - 69.7}{\sqrt{\frac{10.9^2}{16} + \frac{8.2^2}{12} } }$

$t = 0.83$

The given level of significance is

1 - 0.95 = 0.05

The degree of freedom is

df = 16 + 12 - 2 = 26

From the t-table, df = 26 and significance level 0.05,

t = 2.056 (two-tailed)

Conclusion:

We Reject H₀ if t calculated > t tabulated

But in this case,

0.83 is not greater than 2.056

Therefore, We failed to reject H₀

There is no difference between the mean pulse rate of people who do not exercise and the mean pulse rate of people who do exercise.

4 0

2 years ago

Can anyone explain those marked steps in case (iii)

Tema [17]

$\cos\dfrac{3x}2=\cos\left(\dfrac\pi2+\dfrac x2\right)$
$\implies \dfrac{3x}2=2n\pi+\dfrac\pi2+\dfrac x2$

follows from the fact that the cosine function is $2\pi$ -periodic, which means $\cos x=\cos(2\pi+x)$ . Roughly speaking, this is the same as saying that a point on a circle is the same as the point you get by completing a full revolution around the circle (i.e. add $2\pi$ to the original point's angle with respect to the horizontal axis).

If you make another complete revolution (so we're effectively adding $4\pi$ ) we get the same result: $\cos x=\cos(4\pi+x)$ . This is true for any number of complete revolutions, so that this pattern holds for any even multiple of $\pi$ added to the argument. Therefore $\cos x=\cos(2n\pi+x)$ for any integer $n$ .

Next, because $\cos(-x)=\cos x$ , it follows that $\cos x=\cos(2n\pi-x)$ is also true for any integer $n$ . So we have

$\implies \dfrac{3x}2=2n\pi\pm\left(\dfrac\pi2+\dfrac x2\right)$

The rest follows from considering either case and solving for $x$ .