Differentiating Exponential functions In Exercise,find the derivative of the function. See Example 2 and 3.

A plane is traveling 250 mi/hr, how many hours does it take to get to Los Angeles, which is 1300 km away? How many minutes does

inessss [21]

Answer:

Step-by-step explanation:

đổi :250dặm/ giờ= 402.336km/ giờ

thời gian cần để đến Los Angeles: 1300/402.336=3.2311302 giờ=193.867812 phút

6 0

3 years ago

It has been reported that the probability that an individual will develop schizophrenia over their lifetime is 0.004. In a rando

Alinara [238K]

Answer:

Null hypothesis: $p=0.004$

Alternative hypothesis: $p \neq 0.004$

$z=\frac{0.00567 -0.004}{\sqrt{\frac{0.004(1-0.004)}{3000}}}=1.449$

$p_v =2*P(z>1.149)=0.2506$

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion if individuals that developed schizophrenia its not significantly different from 0.004.

Step-by-step explanation:

1) Data given and notation

n=3000 represent the random sample taken

X=17 represent the individuals developed schizophrenia in the sample

$\hat p=\frac{17}{3000}=0.00567$ estimated proportion of individuals developed schizophrenia in the sample

$p_o=0.004$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that true proportion of people who will develop schizophrenia is different from 0.004:

Null hypothesis: $p=0.004$

Alternative hypothesis: $p \neq 0.004$

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

3) Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.00567 -0.004}{\sqrt{\frac{0.004(1-0.004)}{3000}}}=1.449$

4) P value

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a bilateral test the p value would be:

$p_v =2*P(z>1.149)=0.2506$

5) Decision

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion if individuals that developed schizophrenia its not significantly different from 0.004.

4 0

4 years ago

Plz right answer mark u as branliest

neonofarm [45]

Answer:

(120x12)+500

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

If the us dollar changes from 1 = 200 yen to 1 to 100 yen then

Mazyrski [523]