.
Solve the following equations graphically

4. 2x + 3y = 8
x - 2y + 3 =0​

Both arguments are valid.

The validity of the arguments can be determined by using the concept of propositional logic.

This argument can be represented as: (P∨D∨B)∧(D∨C∨B). The form of the argument is p ∧ q, which is a valid form.

This argument can be represented as: P∨C. The form of the argument is p ∨ q, which is also a valid form.

So, both arguments are valid. The validity of an argument depends on the form of the argument and not the specific proposition used. In both cases, the form of the argument is valid, so the argument is also valid.

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Greatest number that divides 300, 560 and 500 is 20 .

Given numbers : 300, 560 and 500

First let’s find prime factors of 300,560 and 500

300 = 2^2 *3^1 *5^2

560= 2^4 * 7^1 *5^1

500 = 2^2 * 5^3

So,

Here highest common power of 2 is 2

Here highest common power of 3 is 0

Here highest common power of 5 is 1

Here highest common power of 7 is 0

Thus HCF (300, 560 and 500) = 2^2 * 5^1 * 3 ^0 * 7 ^0

=4*5*1*1

= 20

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Answer:

x50

Step-by-step explanation:

The equation y = 160x + 80 represents a linear function, where x is the independent variable and y is the dependent variable.

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The proportion of the boards that will be greater than 122 inches is 84%

Given :

length of boards is listed as 10 feet is normally distributed with a mean of 123 inches and a standard deviation of 1 inch.

we are aske to determine what proportion of the boards will be greater than 122 inches = ?

we have :

μ = 123

σ = 122

now, let x denote the number of boards.

⇒ X ∼N (μₓ , σₓ²) X∼N(123,1)

⇒ P(X>122)=P( X- μₓ/ σₓ >  122-123/1)

⇒ P(Z > -1)

= 0.08413

= 84%

hence we get the value as 84%

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​

Mrs Thomas' and Mr Carl's classes both have same lower ratio of A's to B's which expressed in the lowest term as 4:5 or fraction as 4/5

Ratio is used to compare values, and can be expressed as fractions.

Mrs Thomas' class have the ratio of students her students A's to B's as 16:20. And can be expressed as fraction in its lowest term as follows:

16:20 = 16/20

16:20 = 4/5

Mr Carl's class have the ratio of students her students A's to B's as 12:15. And can be expressed as fraction in its lowest term as follows:

12:15 = 12/15

12:15 = 4/5

Therefore, since the lowest form of Mrs Thomas' class and Mr Carl's class ratios of A's to B's are equal, then both class have equivalent lower ratio of A's to B's.

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Answer:4025

Step-by-step explanation:

Answer:

SHADED REGION = 104.06 m2

Step-by-step explanation:

First at all, the square area is:

A1= (22)*(22) = 484 m2

The are for one circle is:

A2 = PI*r2

A2 = (3.14)*(11)2

A2 = 379.94 m2

Then the other circle is the same A2

A2 = A3 = 379.94 m2

You need only half circle for to find the SHADED AREA

(A2)/2 = 189.97 m2

Finally, the SHADED AREA is:

SHADED AREA = A1 -((A2)/2)*2)

SA = 484 m2 - 379.94 m2

SA = 104.06 m2

Answer: m = 13 ;)

Step-by-step explanation:

Answer:

-6m + 12

Step-by-step explanation:

Step 1: Write expression

M + 4 + 2 - 7m + 6

Step 2: Simplify

Answer:

LHS

\(\frac{cos x}{1+ sin x} + \frac{1+ sin x}{cos x} \\=\frac{cos2(x)+1+2sin(x)+sin2(x)}{cos(x)(1+sin(x)} \\= \frac{2+2sin(x)}{cos(x)(1+sin(x))} \\\\=\frac{2 (1 + sin(x))}{cos(x)(1+sin(x))} \\= \frac{2}{cos (x)} \\= 2 sec x\)

= RHS PROVED

hope it helps :)

The identities among the given statements are:

tan x = sin x / cos x

sin x / tan x = cos x

We have,

Let's analyze each of the given statements to determine whether they are identities:

tan x = sin x / cos x:

This is an identity.

It is known as the fundamental identity of trigonometry, as it holds true for all values of x (except when cos x is equal to 0).

tan x cos x = sin x:

This is not an identity.

It is a specific equation that may be true for certain values of x, but it does not hold true for all values of x.

cos x = tan x sin x:

This is not an identity.

Similar to the previous statement, it is a specific equation that may be true for certain values of x, but it does not hold true for all values of x.

sin x / tan x = cos x:

This is an identity.

It can be derived from the fundamental identity by dividing both sides by cos x, resulting in sin x / cos x = cos x / cos x, which simplifies to sin x / tan x = cos x.

Thus,

The identities among the given statements are:

tan x = sin x / cos x

sin x / tan x = cos x

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Answer:

\(5x - 6 = 44 \\ \longrightarrow \: 5x = 44 + 6 \\ \longrightarrow \: 5x = 50 \\ dividing \: both \: sides \: by \: 5 \: \\ \\ \longrightarrow \: \cancel\frac{5x}{5} = \cancel\frac{50}{5} \\ \\ \longrightarrow \:\boxed{ x = 10}\)

hope helpful! :)

Answer:

Step-by-step explanation:

5x - 6 = 44

5x = 50

x = 10

---------------------

check

5 x 10 - 6 = 44     (remember pemdas)

44 = 44

the answer is good

Answer:

Ratio is 9:5 so 14 parts of x. Paul gets 5/14 of x.

Dan gives Paul 32 to make it even split.

So 5/14 ( x) + 32 = 7/14(x)

32 = (7/14 - 5/14)x = 2/14 x = x/7

x= 7(32) = 224

(5/14)224 = 80.

Paul received £80 at first.

Answer:

if this is 4.88/16 then the answer is 0.305.

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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It is assumed here that rats always eat at the same rate, 3 rats eat 3 identical pieces of cheese in 3 hours.

6 rats eat 6 identical pieces of cheese in 6 hours.

Assuming rats eat at the same rate,

3 pieces of cheese last three rats?

It is assumed here that rats always eat at the same rate, 3 rats eat 3 identical pieces of cheese in 3 hours.

Therefore, six rats eat six identical pieces of cheese in six hours and 3 rats eat 3 identical pieces of cheese in 3 hours.

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Answer:n

Step-by-step explanation:

Answer:

4 and 5

Step-by-step explanation:

4^2 = 16

5^2 = 25

12 girls. If there are 40 boys you have to multiply 10 times 4. Using that information you also have to multiply 3 times 4.

This looks like it would be a cylinder

We have two circles on the top and the bottom. If we were to fold these backwards, we would get the circular bases of a cylinder

In the middle, we have a rectangle. If we folded this in a circular fashion around the circlular bases, we would get the long part of a cylinder

Because of this, the shape in its 3D form would be a cylinder

Answer:

400

Step-by-step explanation:

120 + .5x = 80 + .6x

subbtract .5x from each side

120 = 80 + .1x

then subtract 80 from each side

40 = .1x

finally divide by .1 on each side

400 =x

Answer: False

Step-by-step explanation:

Researchers typically report the adjusted R-squared value in addition to the regular R-squared value, not because they lack confidence in the actual R-squared, but because the adjusted R-squared provides additional information about the goodness of fit of a statistical model. The regular R-squared value measures the proportion of the variance in the dependent variable that is explained by the independent variables in the model. However, it can be biased and increase as more predictors are added to the model, even if the additional predictors do not contribute significantly to the prediction.

The adjusted R-squared, on the other hand, takes into account the number of predictors in the model and penalizes the addition of irrelevant predictors. It provides a more conservative measure of the goodness of fit by adjusting for the number of predictors and the sample size. Researchers often use the adjusted R-squared to evaluate and compare different models with varying numbers of predictors or to assess the overall explanatory power of a model while considering its complexity.

In summary, researchers report the adjusted R-squared value to address the limitations of the regular R-squared and to provide a more accurate assessment of the model's goodness of fit.

Answer: 108 sq units

Step-by-step explanation:

Multiply

55 ounces of honey will need to mix with 11 ounces of strawberries.

Question is not complete so lets suppose that,

For each ounce of mix, we use honey five times of a ounces of strawberry.

And then we will find that how many ounces of honey we would to mix with 11 ounces of strawberries.

According to above statement

 one ounce of strawberries = 5 ounce of honey

So, for the mix we will need to add honey which will we equal to five times of strawberries.

If we have 11 ounces of strawberries then,

the amount of honey will be = 5 × 11 = 55 ounces

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the probability of a standard normal random variable being greater than 1.84 is approximately 0.0336.

We can use the central limit theorem to answer this question. The central limit theorem states that if we have a large enough sample size, the distribution of sample means will be approximately normal, even if the population distribution is not normal. The sample mean is given by:

X = μ

where X is the sample mean, μ is the population mean, and n is the sample size.

In this case, we have a sample size of n = 40, and we want to know the probability that the sample mean is greater than 25 minutes. We can standardize the sample mean using the standard error of the mean, which is given by:

SE = σ / sqrt(n)

where σ is the population standard deviation. In this case, σ = 15.4 minutes, so:

SE = 15.4 / sqrt(40) = 2.44 minutes

We can then calculate the z-score for the sample mean:

z = (X - μ) / SE = (25 - 20.5) / 2.44 = 1.84

Using a standard normal distribution table or calculator, we can find that the probability of a standard normal random variable being greater than 1.84 is approximately 0.0336. This means that the probability of a random sample of 40 people having a mean commute time that is greater than 25 minutes is approximately 0.0336, or 3.36%.

In summary, we used the central limit theorem to approximate the distribution of sample means, calculated the standard error of the mean and the z-score for the sample mean, and used a standard normal distribution table or calculator to find the probability of the sample mean being greater than 25 minutes.

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Answer:

17

Step-by-step explanation:

1) Shift the decimals of both 1.36 and 0.08 to the right by 2 decimal places. (This is needed to convert 0.08 to 8, since long division must divide by a whole number.)

1.36 ÷ 8

2) use algorithm method.

3) Solve.

\(17\)

therefor, the answer is 17.

I also added the picture.

hope it helps.

The time that it takes for a ride of 24 miles is given as follows:

1.6 hours.

To obtain the numeric value of a function or of an expression, we replace each instance of the variable in the function or in the expression by the value at which we want to find the desired numeric value.

The distance in this problem is given as the multiplication of the velocity of 15 miles per hour and the time in hours, as follows:

d = 15h.

The variable h can be isolated, thus giving the time in hours for a distance of d miles, as follows:

h = d/15.

The time for a distance of 24 miles is obtained replacing the lone instance of d by 24, as follows:

h = 22

d = 1.6 hours.

The problem asks for the time for a distance of 24 miles.

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Dilation and rotation transformation produces shapes that are not congruent.

Rotations, reflections and translations are known as rigid transformations; this means they do not change the size or shape of a figure, they simply move it. These rigid transformations preserve congruence.

Dilation, however, are not rigid transformations, since they change the size of a shape. Dilation would not change the shape, just the size; the angle measures would be the same, and the ratio of corresponding sides would be equal to the scale factor used in the dilation. This would give us a similar, but not congruent, figure.

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