Which is the best estimate for the quotient 203.12 divided by 48.011 ? My choices are; (A) 4 (B) 40 (C) 400 (D) 4,000

Step-by-step explanation:

This problem is not about percent or relative change, but about absolute change. Its solution is very simple:

Absolute change, or

change = new value - old value

= 56 - 45 = 11 (increase)

Answer:

A

Step-by-step explanation:

Evaluate g(5), then use the value obtained to evaluate f(x) , that is

g(5) = 2(5)² - 4(5) = 2(25) - 20 = 50 - 20 = 30, then

f(30) = 3(30) + 2 = 90 + 2 = 92

Answer:

1, -1

Step-by-step explanation:

a. To find the z score for a given confidence level, you can use the `qnorm()` function in R. The `qnorm()` function takes a probability as an argument and returns the corresponding z score. To find the z score for a 95% confidence level, you can use `qnorm(1-.025)`:

```R
z <- qnorm(1-.025)
```

This will give you the z score for a 95% confidence level, which is approximately 1.96.

b. To create a vector `x` with 50 numbers from a normal distribution with mean 30 and standard deviation 2, you can use the `rnorm()` function:

```R
x <- rnorm(50, mean = 30, sd = 2)
```

To calculate the confidence interval for this data, you can use the formula:

```R
CI <- mean(x) + c(-1, 1) * z * sd(x) / sqrt(length(x))
```

This will give you the lower and upper bounds of the 95% confidence interval. You can check whether the interval covers the true mean of 30 by seeing if 30 is between the lower and upper bounds:

```R
lower <- CI[1]
upper <- CI[2]
if (lower <= 30 && upper >= 30) {
 print("The interval covers the true mean.")
} else {
 print("The interval does not cover the true mean.")
}
```

c. To repeat the above experiment 200 times and calculate the percentage of intervals that cover the true mean, you can use a for loop:

```R
count <- 0
for (i in 1:200) {
 x <- rnorm(50, mean = 30, sd = 2)
 CI <- mean(x) + c(-1, 1) * z * sd(x) / sqrt(length(x))
 lower <- CI[1]
 upper <- CI[2]
 if (lower <= 30 && upper >= 30) {
   count <- count + 1
 }
}
percentage <- count / 200
```

This will give you the percentage of intervals that cover the true mean.

d. To write a function that takes a confidence level as an input and returns the percentage of intervals that cover the true mean, you can use the following code:

```R
calculate_percentage <- function(CL) {
 z <- qnorm(1-(1-CL)/2)
 count <- 0
 for (i in 1:200) {
   x <- rnorm(50, mean = 30, sd = 2)
   CI <- mean(x) + c(-1, 1) * z * sd(x) / sqrt(length(x))
   lower <- CI[1]
   upper <- CI[2]
   if (lower <= 30 && upper >= 30) {
     count <- count + 1
   }
 }
 percentage <- count / 200
 return(percentage)
}
```

You can then use this function to create a 5 by 2 matrix with one column showing the theoretical CL and the other showing the empirical coverage probability:

```R
CL <- c(.8, .85, .9, .95, .99)
percentage <- sapply(CL, calculate_percentage)
matrix <- cbind(CL, percentage)
```

This will give you a matrix with the theoretical CL in the first column and the empirical coverage probability in the second column.

Know more about z score here:

brainly.com/question/15016913

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