Write the expression as the sine, cosine, or tangent of an angle.sin 9x cos x - cos 9x sin x

Answer:

there are 1.54 months in 47 days

Step-by-step explanation:

47 days is equal to 1.52 months. This is also 67680 minutes, 1128 hours, 47 days, 5.88 work days, 6.71 weeks, 1.52 months, And is 12.88% through the year. Converting days is used mostly to track time for different contexts. For example, days might make more sense for personal calendars while work days, weeks, and percent through the year could make more sense for financial or corporate calendars. 47 days could be calculated in multiple ways, and this page will walk you through step by step to convert 47 days to 1.52 months.

The expression 81⁽¹/²⁾ simplifies to 9.

To simplify the expression 81¹/², we can evaluate the squareroot of 81.

The square root of a number x is a value that, when multiplied by itself, gives x. In this case, we're looking for the number that, when squared, equals 81.

The square root of 81 is 9 since 9 x 9 = 81.

Therefore, 81⁽¹/²⁾ simplifies to 9.

In terms of steps, we can represent the process as follows:

1. Recognize that 81⁽¹/²⁾ represents the square root of 81.

2. Evaluate the squareroot of 81, which is 9.

3. Thus, 81^(1/2) simplifies to 9.

By simplifying the expression, we find that 81⁽¹/²⁾is equal to 9.

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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.

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