Answer:
5 is NOT a solution to the equation.
t = 6
Step-by-step explanation:
Step 1: Write out equation
2t - 1 = 11
Step 2: Add 1 to both sides
2t = 12
Step 3: Divide both sides by 2
t = 6
Answer:
c
Step-by-step explanation:
hope this helps you out
This is a change in both supply and demand. Due to farmers raising fewer cows the supply of cows goes down. Because there is more of a need of cows due to the shortage the demand or need for them goes up. Due to the limited amount of cows and the high demand prices will go up.
Answer:
A.102%
Step-by-step explanation:
Let cost price of painting=$100
In first year price increased 20%
Then , the price=
$120
In second year
Price decreased 15%
Then , the price of painting=![120-120(0.15)](https://tex.z-dn.net/?f=120-120%280.15%29)
The price of painting=$102
Percent =![\frac{final\;price}{Initial\;price}\times 100](https://tex.z-dn.net/?f=%5Cfrac%7Bfinal%5C%3Bprice%7D%7BInitial%5C%3Bprice%7D%5Ctimes%20100)
By using this formula
Then, we get
Percent of the original price=![\frac{102}{100}\times 100](https://tex.z-dn.net/?f=%5Cfrac%7B102%7D%7B100%7D%5Ctimes%20100)
Percent of the original price=102%
Option A is true.
Answer:
![Mean=p=0.75](https://tex.z-dn.net/?f=Mean%3Dp%3D0.75)
![sd=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.75(1-0.75)}{450}}=0.0204](https://tex.z-dn.net/?f=sd%3D%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D%3D%5Csqrt%7B%5Cfrac%7B0.75%281-0.75%29%7D%7B450%7D%7D%3D0.0204)
Step-by-step explanation:
1) Data given
represent the population proportion of complaints settled for new car dealers
represent the sample of complaints involving new car dealers.
2) Find the distribution of ![\hat p](https://tex.z-dn.net/?f=%5Chat%20p)
First we can begin with the expected value
and that represent the mean
Now we can find the variance for ![\hat p](https://tex.z-dn.net/?f=%5Chat%20p)
When we use a proportion p, when we draw n items each from a Bernoulli distribution. The variance of each Xi distribution is p(1−p) and hence the standard error is p(1−p)/n. for this reason the variance for
is given by:
![Var(\hat p)= \frac{p(1-p)}{n}](https://tex.z-dn.net/?f=Var%28%5Chat%20p%29%3D%20%5Cfrac%7Bp%281-p%29%7D%7Bn%7D)
So then the deviation would be given by:
![Sd(\hat p)=\sqrt{\frac{p(1-p)}{n}}](https://tex.z-dn.net/?f=Sd%28%5Chat%20p%29%3D%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D)
The sample distribution of the sample proportion
is normal, so then we have this:
![\hat p \sim N(p,\sqrt{\frac{p(1-p)}{n}})](https://tex.z-dn.net/?f=%5Chat%20p%20%5Csim%20N%28p%2C%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D%29)
3) Calculating the mean and standard deviation
We can replace the values given in order to find the mean and deviation:
![Mean=p=0.75](https://tex.z-dn.net/?f=Mean%3Dp%3D0.75)
![sd=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.75(1-0.75)}{450}}=0.0204](https://tex.z-dn.net/?f=sd%3D%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D%3D%5Csqrt%7B%5Cfrac%7B0.75%281-0.75%29%7D%7B450%7D%7D%3D0.0204)