May i get help please?

Six students measure the acceleration (in meters per second per second) of object in free fall. The measured values are shown. T

Sunny_sXe [5.5K]

The measured data is
x=[10.56, 9.52, 9.73, 9.80, 9.78, 10.91].

Calculate the mean.
m = (10.56+9.52+9.73+9.80+9.78+10.91)/6 = 10.05

Calculate the deviations from the mean.
k = x - m
= [0.51, -0.53, -0.32, -0.25, -0.27, 0.86]

Calculate the absolute deviation from the mean.
$d= \frac{1}{6} | (0.51-0.53-0.32-0.25-0.27+0.86) | = |1.184\times10^{-15}| \approx 0$

Answer: 0

6 0

4 years ago

The risk of a child developing cancer is approx 3 in 1500. If there are approx 11,721,722 children, how many have cancer?

Maslowich

Answer:

Approximately 23,433 children will have cancer.

Step-by-step explanation:

3/1500 can be simplified to 1/500, which can also be written as 0.002. To find the number of children who have cancer, we do 11,721,722 * 0.002, which gives us 23,433.444 which we can round to 23,433.

4 0

3 years ago

A student ran out of time on a multiple choice exam and randomly guessed the answers for 2 problems. Each problem had 5 answer c

frozen [14]

So if there are 5 choices and only one is correct, there is a 1/5 chance of being correct. since there was 2 questions, you need to multiply 1/5 by 1/5 to get 2/25 of getting both questions correct.

i hope this helps!!

3 0

3 years ago

A manufacturer of potato chips would like to know whether its bag filling machine works correctly at the 420 gram setting. It is

anastassius [24]

Answer:

$t=\frac{424-420}{\frac{26}{\sqrt{61}}}=1.202$

$p_v =2*P(t_{(60)}>1.202)=0.234$

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 reject the null hypothesis, so we can conclude that the true mean is NOT different from 420. So the specification is satisfied.

Step-by-step explanation:

Data given and notation

$\bar X=424$ represent the sample mean

$s=26$ represent the sample standard deviation

$n=61$ sample size

$\mu_o =420$ represent the value that we want to test

$\alpha=0.01$ 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 true mean is different from 420 or not, the system of hypothesis would be:

Null hypothesis: $\mu = 420$

Alternative hypothesis: $\mu \neq 420$

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{424-420}{\frac{26}{\sqrt{61}}}=1.202$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=61-1=60$

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

$p_v =2*P(t_{(60)}>1.202)=0.234$

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 reject the null hypothesis, so we can conclude that the true mean is NOT different from 420. So the specification is satisfied.

4 0

3 years ago

I really think it’s B ...Help?

kvv77 [185]

The correct answer is c y-5=4(x-7)

7 0

3 years ago