Simplify 5x + 3y - 2x + 4y

Answers

Answer 1

Answer: 5x+3y-2x+4y= 5x-2x+4y+3y= 3x+7y

Answer 2

Answer: 5x - 2x = 3x
3y + 4y = 7y
Answer: 3x + 7y

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Can you help with the question

Answers

Step-by-step explanation:

See image

Samir buys x cans of soda at 30 cents each and (x+4) cans of soda at 35 cents each. The total cost was $3.35. Find x

Answers

An equation is formed of two equal expressions. The value of x is 3.23.

What is an equation?

An equation is formed when two equal expressions are equated together with the help of an equal sign '='.

The cost of x cans can be written as $=x * 0.30$

The cost of (x+4) cans can be written as,

$(x+4) * 0.35\n= 0.35 x+ 1.4$

Now, the total cost of the cans is given to us like $3.35, therefore, the value of x can be written as,

$(0.30x) + 0.35x+ 1.4 = 3.35\n\n0.65x = 3.5-1.4\n\nx = (2.1)/(0.65)\n\nx = 3.23$

Hence, the value of x is 3.23.

Learn more about Equation:

brainly.com/question/2263981

0.30x is the cost of the first type of soda. 0.35(x+4) is the cost of the second type of soda. Put these into an equation looking like:

Cost of Soda A + Cost of Soda B = Total Cost
0.30x + 0.35(x+4)=3.35 --> Expand 0.35(x+4)
0.30x + 0.35x + 1.4 = 3.35 --> Add the x's and subtract 1.4
0.65x = 1.95 --> Divide by 0.65
x = 3

Samir buys 3 cans of soda at 30 cents each and 7 (3+4) cans of soda at 35 cents each. Samir buys a total of 10 cans of soda.

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Answers

Answer:

:) /

Step-by-step explanation:,,

Answer:

The answer is B:subtract 1,add x,divide by 4

The other answer is D:add x,subtract 1,divide 4

Step-by-step explanation:

when you subtract 1 you isolate 3x so the equation looks like 3x=-x+3

the next step is to add x to both sides so you isolate the x to one side after doing that your equation should look like 4x=3,after this you divide both sides to the solution x= $(3)/(4)$

The answer is also D because if you add x to both sides u isolate x to the left side so the equation will look like 1+4x=4 the next step will be to subtract one from each side and after doing this the equation should look like 4x=3 which should look familiar because its the same exact solution as B was so the answer to this choice is x= $(3)/(4)$

With that evidence I can reasonably say that answer to the question is B,and D

Solve each of thr following sytem of equations by substitution!!

4x+3y=37
y=x-4

Answers

x = 7, y = 3
yeeeeeeee

Ou have learned that given a sample of size n from a normal distribution, the CL=95% confidence interval for the mean can be calculated by Sample mean +/- z((1-CL)/2)*Sample std/sqrt(n). Where z((1-cl)/2)=z(.025) is the z score.a. help(qnorm) function. Use qnorm(1-.025) to find z(.025).
b. Create a vector x by generating n=50 numbers from N(mean=30,sd=2) distribution. Calculate the confidence interval from this data using the CI formula. Check whether the interval covers the true mean=30 or not.
c. Repeat the above experiments for 200 times to obtain 200 such intervals. Calculate the percentage of intervals that cover the true mean=30. This is the empirical coverage probability. In theory, it should be very close to your CL.
d. Write a function using CL as an input argument, and the percentage calculated from question c as an output. 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, for CL=.8, .85, .9, .95,.99.

Answers

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.