Guys help me on this one please…

ankoles [38]

Because without it the economy would fail :)

4 0

3 years ago

Part b compare the bond enthalpies of the nitrogen-nitrogen single and triple bonds. based on this information, what is the aver

s2008m [1.1K]

C-C bonds = 348 C=C bonds = 614 C(triple) bonds = 839
All in KJ/Mol
just do 839-614=225 and 614-348=266
After a pi bond is added on, these two numbers are the KJ/Mol it takes to separate the C-C bond,
To get the average: (225+266)/2 = 245.5Kj/Mol
You might want to round your answer to 246KJ/Mol

5 0

3 years ago

Given the equation g(x)= – x2–5x + 12 find g(9)

Natasha2012 [34]

Answer:

g(9) = 48

Step-by-step explanation:

g(x)= – x2–5x + 12

I'll assume this is

g(x)= – x^2–5x + 12

find g(9)

g(9)= – 9^2–5(9) + 12

g(9)= 81–45 + 12

g(9) = 48

8 0

3 years ago

The mean number of hours per day spent watching television, according to a national survey, is 3.5 hours, with a standard deviat

aleksley [76]

Answer:

4.5 hour and 2 hour

Step-by-step explanation:

Given: The mean number of hours per day spent watching television, according to a national survey, is 3.5 hours, with a standard deviation of two hours.

To Find: If each time was increased by one hour, what would be the new mean and standard deviation.

Solution:

let the total numbers entries of hours in survey be = $\text{N}$

each entry in survey be = $\text{x}_{i}$

mean of survey is = $\mu$ = $3.5$ $\text{hours}$

standard deviation is = $\sigma$ = $2$ $\text{hours}$

if each entry in survey is increased by one hour then,

each new entry in survey is = $\text{x}_{i}+1$

the new mean is $\mu_{new}$ = $\frac{\text{sum of all hours}}{total number of entries}$

$\frac{\text{x}_{1} +1+\text{x}_{2}+1 +....+\text{x}_{N}+1 }{N}$

$\frac{(\text{x}_{1} +\text{x}_{2} +....+\text{x}_{N})+\text{N}\times1 }{N}$

$\frac{(\text{x}_{1} +\text{x}_{2} +....+\text{x}_{N})}{N}$ + $\frac{\text{N}\times1 }{N}$

$\mu_{new}$ = $\mu$ + 1= $4.5$ $\text{hours}$

now,

standard deviation is $\sigma_{new}=$ $\sqrt{\sum_{1}^{N}\frac{(\text{x}_{inew}-\mu_{new})^{2}}{N}}$

$\text{x}_{inew}= \text{x}_{i}+1$

$\mu_{new}=\mu+1$

putting values,

$\sqrt{\sum_{1}^{N}\frac{(\text{x}_{i}+1-\mu-1)^{2}}{N}}$

$\sqrt{\sum_{1}^{N}\frac{(\text{x}_{i}-\mu)^{2}}{N}}$ = $\sigma$

$\sigma_{new}$ =2

new mean and standard deviation are $4.5$ and $2$ $\text{hours}$

5 0

3 years ago

Suppose that 500 parts are tested in manufacturing and 10 are rejected.

alexdok [17]

Answer:

a) $z=\frac{0.02 -0.03}{\sqrt{\frac{0.03(1-0.03)}{500}}}=-1.31$

$p_v =P(Z$

If we compare the p value obtained 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 of rejected items is less than 0.03.

b) We can use a 95 percent upper confidence interval.

On this case we want a interval on this form : $(-\infty,\hat p +z_{\alpha}\sqrt{\frac{\hat p (1-\hat p)}{n}})$

So the critical value would be on this case $Z_{\alpha}=1.64$ and we can use the following excel code to find it: "=NORM.INV(1-0.05,0,1)"

We found the upper limit like this:

$0.02+1.64\sqrt{\frac{0.02 (1-0.02)}{500}}=0.03026$

And the interval would be: $(-\infty,0.03026)$

And since our value (0.02) is contained in the interval We fail to reject the hypothesis that p=0.03

Step-by-step explanation:

Part a

Data given and notation

n=500 represent the random sample taken

X=10 represent the number of objects rejected

$\hat p=\frac{10}{500}=0.02$ estimated proportion of objects rejected

$p_o=0.03$ 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)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that 70% of adults say that it is morally wrong to not report all income on tax returns.:

Null hypothesis: $p=0.03$

Alternative hypothesis: $p < 0.03$

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 $p_o$ .

Calculate the statistic

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

$z=\frac{0.02 -0.03}{\sqrt{\frac{0.03(1-0.03)}{500}}}=-1.31$

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

Since is a one tailed left test the p value would be:

$p_v =P(Z$

If we compare the p value obtained 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 of rejected items is less than 0.03.

Part b

We can use a 95 percent upper confidence interval.

On this case we want a interval on this form : $(-\infty,\hat p +z_{\alpha}\sqrt{\frac{\hat p (1-\hat p)}{n}})$

So the critical value would be on this case $Z_{\alpha}=1.64$ and we can use the following excel code to find it: "=NORM.INV(1-0.05,0,1)"

We found the upper limit like this:

$0.02+1.64\sqrt{\frac{0.02 (1-0.02)}{500}}=0.03026$

And the interval would be: $(-\infty,0.03026)$

And since our value (0.02) is contained in the interval We fail to reject the hypothesis that p=0.03

3 0

3 years ago