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2 years ago

What is the average rate of change of the function on the interval from x = 0 to x = 5?

Mathematics

2 answers:

bixtya [17]2 years ago

8 0

The average rate is 3^x/2

damaskus [11]2 years ago

6 0

Answer:

Average rate of change = 24.2

Step-by-step explanation:

$f(x)=\frac{1}{2}3^x$

Average rate of change formula is on interval a to b is

$\frac{f(b)-f(a)}{b-a}$

Given interval is x=0 to x=5

a=0 and b=5

Average rate = $\frac{f(5)-f(0)}{5-0}$

$f(0)=\frac{1}{2}3^0=\fracc{1}{2}=0.5$

$f(5)=\frac{1}{2}3^5=\frac{3^5}(2}=121.5$

Average rate = $\frac{f(5)-f(0)}{5-0}$

Average rate = $\frac{121.5-0.5}{5}=24.2$

Average rate of change = 24.2

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What is the exact value of tan 30° ? Enter your answer, as a simplified fraction, in the box.

Romashka [77]

Hmmm if you don't have a Unit Circle, this is a good time to get one, many you can find online. Anyhow, check your unit circle for cos(30°) and sin(30°).

$\bf tan(\theta)=\cfrac{sin(\theta)}{cos(\theta)}\qquad \qquad tan(30^o)=\cfrac{sin(30^o)}{cos(30^o)}
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tan(30^o)=\cfrac{\quad \frac{1}{2}\quad }{\frac{\sqrt{3}}{2}}\implies tan(30^o)=\cfrac{1}{2}\cdot \cfrac{2}{\sqrt{3}}\implies tan(30^o)=\cfrac{1}{\sqrt{3}}
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\stackrel{\textit{and now if we rationalize the denominator}}{\cfrac{1}{\sqrt{3}}\cdot \cfrac{\sqrt{3}}{\sqrt{3}}\implies \cfrac{\sqrt{3}}{(\sqrt{3})^2}\implies \cfrac{\sqrt{3}}{3}}$

8 0

2 years ago

Read 2 more answers

In a random sample of 75 American women age 18 to 30, 26 agreed with the statement that a woman should have the right to a legal

ddd [48]

Answer:

a) $z=\frac{0.347-0.328}{\sqrt{0.338(1-0.338)(\frac{1}{75}+\frac{1}{64})}}=0.236$

$p_v =2*P(Z>0.236)=0.813$

If we compare the p value and using any significance level for example $\alpha=0.01$ always $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say that we don't have significant differences between the two proportions.

b) We are confident at 99% that the difference between the two proportions is between $-0.188 \leq p_B -p_A \leq 0.226$

Step-by-step explanation:

Previous concepts and data given

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$p_A$ represent the real population proportion of women age 18 to 30 agreed with the statement that a woman should have the right to a legal abortion for any reason

$\hat p_A =\frac{26}{75}=0.347$ represent the estimated proportion of women age 18 to 30 agreed with the statement that a woman should have the right to a legal abortion for any reason

$n_A=75$ is the sample size for A

$p_B$ represent the real population proportion for women age 58 to 70 agreed with the statement that a woman should have the right to a legal abortion for any reason

$\hat p_B =\frac{21}{64}=0.328$ represent the estimated proportion of women age 58 to 70 agreed with the statement that a woman should have the right to a legal abortion for any reason

$n_B=64$ is the sample size required for B

$z$ represent the critical value for the margin of error and for the statisitc

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Part a

We need to conduct a hypothesis in order to check if the proportion are equal, the system of hypothesis would be:

Null hypothesis: $p_{A} = p_{B}$

Alternative hypothesis: $p_{A} \neq p_{B}$

We need to apply a z test to compare proportions, and the statistic is given by:

$z=\frac{p_{A}-p_{B}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{A}}+\frac{1}{n_{B}})}}$ (1)

Where $\hat p=\frac{X_{A}+X_{B}}{n_{A}+n_{B}}=\frac{26+21}{75+64}=0.338$

Calculate the statistic

Replacing in formula (1) the values obtained we got this:

$z=\frac{0.347-0.328}{\sqrt{0.338(1-0.338)(\frac{1}{75}+\frac{1}{64})}}=0.236$

Statistical decision

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

$p_v =2*P(Z>0.236)=0.813$

Part b

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}$

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.