A mathematician is working with a programmer to write a program to solve a problem using high level mathematics. The mathematician asks the programmer to help her determine the efficiency of the algorithm. How can the efficiency be determined?

Answer: 12 minutes

Step-by-step explanation:

This is a standard Work Formula question (2 or more 'entities' that work on a task together). When there are just 2 entities and there are no 'twists' to the question, we can use the Work Formula to get to the correct answer.

Work = (A)(B)/(A+B) where A and B are the individual times needed to complete the task.

We're told that two hoses take 20 minutes and 30 minutes, respectively, to fill a pool. We're asked how long it takes the two hoses, working together, to fill the pool.

(20)(30)/(20+30) = 600/50 = 12 minutes to fill the pool.

Answer:

Option C is correct that is h(x)=x-3

Step-by-step explanation:

We have been given a function:

Hence,

Therefore,

Hence, Option C is correct that is h(x)=x-3

Answer:

Option A is correct

The density approaches infinity.

Step-by-step explanation:

Given the equation:

              ....[1]

where,

D is density of a particular substance .

v is the volume of the substance.

We have to find What happens to the density as the volume approaches 0.

As the volume approaches to 0

⇒

[1] ⇒

⇒The density approaches infinity.

Therefore, the density as the volume approaches 0 is,  infinity.

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

#SPJ11

Answer:

23

Step-by-step explanation:

(f + g)(3) = f(3) + g(3)

f(3) = 3(3) - 3 = 9 - 3 = 6

g(3) = 4(3) + 5 = 12 + 5 = 17

Then

(f + g)(3) = 6 + 17 = 23