Solve for C<br><br> C<br> — - 4.3 = 11.5 <br> 2

Luiza is jumping on a trampoline. Ht models her distance above the ground (in m) t seconds after she starts

Damm [24]

Answer:

t=2.08 seconds.

Step-by-step explanation:

Well, in this example, H(t)=-0.6cos(2pi/2.5)t+1.5 should be equal to 1.2. If calculated, -0.6cos(0.8pi)t=1.2-1.5 which is equal to -0.6cos(0.8pi)t=-0.3, then cos(0.8pi)t=0.5. The value of cosine in terms of radians when it is equal to 0.5 is pi/3. So, cos(0.8pi)t=cos(pi/3). If simplified, (0.8pi)*t=5pi/3. pi's are cancelled out and t is calculated as 2.08333... If rounded to the nearest hundredth it is 2.08.

4 0

2 years ago

Why shouldn't you use double negatives? It's a riddle for my Math Class

IRINA_888 [86]

A negative times a negative is a positive. I am not sure if that's the question, but hopefully that helps if it was.

3 0

3 years ago

Read 2 more answers

In 2006, the population of Tewksbury, Rhode Island was 25,000, and it was growing at an annual rate of 2.2%. What is the growth

Shtirlitz [24]

Tewksbury Population In 2011

=
$25000 {( {1.022)}^{2011 - 2006} }$
= 27,874

7 0

3 years ago

Read 2 more answers

Can someone plz help me with #1-8?

slamgirl [31]

Answer:9

Step-by-step explanation:

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

Solve for C C — - 4.3 = 11.5 2