The diagram shows a circle inside a square 16cm Work out the area of the circle

Find the nth term of this sequence 5 10 15 20 25 ...

densk [106]

The NTH term is 5n..

6 0

3 years ago

A sample of 900900 computer chips revealed that 76v% of the chips do not fail in the first 10001000 hours of their use. The comp

Sergio039 [100]

Answer:

Null hypothesis: $p=0.78$

Alternative hypothesis: $p \neq 0.78$

$z=\frac{0.76 -0.78}{\sqrt{\frac{0.78(1-0.78)}{900}}}=-1.448$

$p_v =2*P(Z$

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ 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 of chip that do not fail in the first 1000 hours is not significantly different from 0.78 or 78%.

Step-by-step explanation:

1) Data given and notation

n=900 represent the random sample taken

$\hat p=0.76$ estimated proportion of chips do not fail in the first 1000 hours

$p_o=0.78$ is the value that we want to test

$\alpha$ represent the significance level

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 the true proportion is 0.78:

Null hypothesis: $p=0.78$

Alternative hypothesis: $p \neq 0.78$

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:

$z=\frac{0.76 -0.78}{\sqrt{\frac{0.78(1-0.78)}{900}}}=-1.448$

4) Statistical decision

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 assumed for this case is $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a bilateral test the p value would be:

$p_v =2*P(Z$

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ 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 of chip that do not fail in the first 1000 hours is not significantly different from 0.78 or 78%.

7 0

3 years ago

Five students volunteered to carry boxes.There are 62 boxes.Is it possible for each student to carry the same number of boxes an

Anni [7]

If you divide 62 boxes by 5 students, you will get 12.4. You cannot divide one box by 5 for each student. In this case, your answer is no. The students would not get an even amount of boxes each.

5 0

4 years ago

Read 2 more answers

Someone help me plssssss

zlopas [31]

Answer:

(3,-4)

Step-by-step explanation:

3 0

4 years ago

Read 2 more answers

Finding second Derivatives In Exercise,find the second derivate.<br> f(x) = (3 + 2x)e - 3x

Talja [164]

Answer:

the question is incomplete, the complete question is "Finding second Derivatives In Exercise,find the second derivate.

$f(x)=(3+2x)e^{-3x}$ "

answer: $\frac{df(x)^{2}}{dx^{2}}=(15+18x)e^{-3x}\\$ ,

Step-by-step explanation:

To determine the second derivative, we differentiate twice.

for the first differentiation, we use the product rule approach. i.e

$f(x)=u(x)v(x)\\\frac{df(x)}{dx}=u(x)\frac{dv(x)}{dx}+ v(x)\frac{du(x)}{dx}\\$

from $f(x)=(3+2x)e^{-3x}$ if w assign

u(x)=(3+2x) and the derivative, $\frac{du(x)}{dx}=2$

also $v(x)=e^{-3x}$ and the derivative $\frac{dv(x)}{dx}=-3e^{-3x}$ .

If we substitute values we arrive at

$\frac{df(x)}{dx}=(3+2x)(-3e^{-3x})+2e^{-3x}\\\frac{df(x)}{dx}=(-9-6x)e^{-3x}+2e^{-3x}\\\frac{df(x)}{dx}=(-9-6x+2)e^{-3x}\\\frac{df(x)}{dx}=-(7+6x)e^{-3x}\\$ ,

Now to determine the second derivative we use the product rule again

this time, u(x)=(7+6x) and the derivative, $\frac{du(x)}{dx}=6$

also $v(x)=e^{-3x}$ and the derivative $\frac{dv(x)}{dx}=-3e^{-3x}$ .

If we substitute values we arrive at

$\frac{df(x)^{2}}{dx^{2}}=-((7+6x)(-3e^{-3x})+6e^{-3x})\\\frac{df(x)^{2}}{dx^{2}}=-((-21-18x)e^{-3x}+6e^{-3x})\\\frac{df(x)^{2}}{dx^{2}}=-((-21-18x+6)e^{-3x})\\\frac{df(x)^{2}}{dx^{2}}=(15+18x)e^{-3x}\\$ ,

7 0

3 years ago