During practice, the Northwood football team drinks water from a cylindrical cooler that has a radius of 6 inches and a height of 20 inches. Players use conical paper cups, as shown below. A cone has a diameter of 4.4 inches and a height of 5.5 inches. If the water cooler is filled completely, can each of the 38 players have two full paper cups of water during practice? Explain.

The function that is based on the translation is g(x) = (x + 5)^2

How to determine the translation?

The function f(x) is given as:

f(x) = (x + 7)^2

The function f(x) is translated 2 units right to get function g(x)

This means that:

g(x) = f(x - 2)

So, we have:

f(x - 2) = (x -2 + 7)^2

Evaluate

f(x - 2) = (x + 5)^2

This gives

g(x) = (x + 5)^2

Hence, the function that is based on the translation is g(x) = (x + 5)^2

Read more about translation at:

brainly.com/question/8959437

#SPJ5

Your Answer: G(x) = (x − 5)²

I just finished my semester exam, and this was one of the questions. I guessed on this question because I had no clue of what the answer was, and lucky me, I got the question right :D

Hope this helps y'all with your test :)

Answer:

Bianca's height = 42 inches

Step-by-step explanation:

Let x be the Bianca height.

Given:

Meredith height = 60 inches

We need to find the Bianca height.

Solution:

From the given statement the Meredith's height is of Bianca's height plus 32 inches, so the equation is.

Meredith's height =

Substitute Meredith's height in above equation.

Now we solve the above equation for x.

By cross multiplication.

28 divided by 2.

Therefore, the height of the Bianca is 42 inches.

The line AB touching the circle at point B in the considered diagram is not tangent to the circle O.

What is Pythagoras Theorem?

If ABC is a triangle with AC as the hypotenuse and angle B with 90 degrees then we have:

where |AB| = length of line segment AB. (AB and BC are rest of the two sides of that triangle ABC, AC being the hypotenuse).

How are radius and tangent to a circle related?

There is a theorem in mathematics that:

If there is a circle O with tangent line L intersecting the circle at point A, then the radius OA is perpendicular to the line L.

So, if AB is a tangent, then ∠ABO = 90° and therefore satisfies Pythagoras theorem.

Assuming AB is tangent, then ABO is right angled we should get:

This statement is false, and therefore, so as our assumption is false that ABis tangent to circle O. Thus, AB is not tangent to circle O.

(so it might be that even if AB looks like touching at one point the circle O, but AB might be intersecting the circle at two points, or not touching it at all)

Thus, the line AB touching the circle at point B in the considered diagram is not tangent to the circle O.

Learn more about tangent to a circle here:

brainly.com/question/7942024

Answer:

not tangent

Step-by-step explanation:

two reasons, first

Triangle AOB is not a right triangle

line AB intersects the circle O at two points.

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