Clark's rule is a formula used to determine pediatric doses of over the counter medicine.
The formula says
The weight of a child divided by 150 multiplied by the adult dose is the same as the child dose
if an adult dose of acetaminophen is 1,000 mg and a child weighs less than 90 pounds, what is
the recommended range of doses?
A. x≤ 600
B. x ≥ 600
C. 500 D. 500 ≤x≤ 600

Poppy's sister used approximately 3.14 inches more wire per wall hanging than Poppy.

To find the amount of wire used by both Poppy and her sister, we'll calculate the circumference of their respective wall hangings using the formula:

C = 2πr

Since the diameter is twice the radius, we can find her sister's radius as follows:

Poppy's radius = 6.25 inches

Poppy's diameter = 2 * 6.25 = 12.5 inches

Sister's diameter = Poppy's diameter + 1 = 12.5 + 1 = 13.5 inches

Sister's radius = 13.5 / 2 = 6.75 inches

Now, we'll calculate the circumferences using the given value for π:

Poppy's circumference = 2 x 3.14 x 6.25 ≈ 39.25 inches

Sister's circumference = 2 x 3.14 x 6.75 ≈ 42.39 inches

Difference = Sister's circumference - Poppy's circumference

Difference = 42.39 - 39.25 = 3.14 inches

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Answer:

Let's use a Venn diagram to solve this problem.

First, let's label the three regions of the Venn diagram: Computer only (C), Optional Maths only (M), and both Computer and Optional Maths (C ∩ M). We can also label the region outside the circles as neither Computer nor Optional Maths (N).

We know that 30% of the students like Computer only, which means that the percentage of students in region C is 30%. Similarly, 25% of the students like both Computer and Optional Maths, so the percentage of students in region C ∩ M is 25%.

We are also given that 5% of the students don't like either subject, so the percentage of students in region N is 5%.

Finally, we are told that 390 students like Optional Maths, which includes the students in regions M and C ∩ M. We don't know the percentage of students in region M, but we do know that the percentage of students in region C ∩ M is 25%.

Using this information, we can set up an equation to solve for the total number of students:

C + M + C ∩ M + N = 100%

Substituting the percentages we know, we get:

30% + M + 25% + 5% = 100%

Simplifying the equation, we get:

M = 40%

This means that 40% of the students like Optional Maths only, which is the percentage of students in region M.

Now we can use the fact that 390 students like Optional Maths to solve for the total number of students:

M + C ∩ M = 390

0.4T + 0.25T = 390

0.65T = 390

T = 600

Therefore, the total number of students is 600.

Answer:

5/1

Step-by-step explanation:

Answer:

5/1

Step-by-step explanation:

Answer: 28

Step-by-step explanation:

To find the number, we need to use the fact that 4/7 of a number is 84 to set up an equation that we can solve. Since 4/7 of a number is 84, we can write the following equation:

(4/7) * x = 84

To solve this equation, we need to first get rid of the fraction by multiplying both sides of the equation by 7/4:

(4/7) * x * (7/4) = 84 * (7/4)

This gives us the equation:

(4/7) * (7/4) * x = 84 * (7/4)

Since (4/7) * (7/4) = 1, we can simplify the equation as follows:

1 * x = 84 * (7/4)

We can then simplify the right side of the equation by multiplying the numerator and denominator of the fraction by 4:

1 * x = 84 * (7/4) = 84 * (28/28) = 28

Finally, we can solve for x by dividing both sides of the equation by 1:

1 * x / 1 = 28 / 1

This gives us the final equation:

x = 28

Therefore, the number is 28.

The PDF of the random variable \(\(Z = \frac{Y}{X}\) is \(f_Z(z) = z\) for \(0 \leq z \leq 1\) and \(f_Z(z) = 0\)\) otherwise.

To find the probability density function (PDF) of the random variable \(\(Z = \frac{Y}{X}\)\), where \(X\) and \(Y\) are independent random variables uniformly distributed on the interval \([0,1]\), we can use the method of transformations.

Let's denote the PDF of \\(\(X\) as \(f_X(x)\) and the PDF of \(Y\) as \(f_Y(y)\).\)

Since X and Y are independent and uniformly distributed on [0,1], their PDFs are constant within that interval and zero outside of it. Therefore, we have:

\(\(f_X(x) = \begin{cases} 1, & 0 \leq x \leq 1 \\ 0, & \text{otherwise} \end{cases}\)\)

\(\(f_Y(y) = \begin{cases} 1, & 0 \leq y \leq 1 \\ 0, & \text{otherwise} \end{cases}\)\)

To find the PDF of Z, we need to determine the cumulative distribution function (CDF) of Z and then differentiate it to obtain the PDF.

The cumulative distribution function (CDF) of Z is given by:

\(\(F_Z(z) = P(Z \leq z)\)\)

To find \(\(P(Z \leq z)\)\), we can consider the range of values that Z can take. Sinc\(\(0 \leq X \leq 1\) and \(0 \leq Y \leq 1\), the range of \(Z\) is \(0 \leq Z \leq \frac{1}{0} = \infty\) (assuming \(X \neq 0\)).\)

For a given z, Z can take values less than or equal to z if and only if\(\(Y \leq zX\).\)

Therefore, we can express \(\(P(Z \leq z)\)\) as an integral:

\(\(P(Z \leq z) = \int_{0}^{1} \int_{0}^{zx} f_Y(y) f_X(x) \, dy \, dx\)\)

Substituting the PDFs of \(X\) and \(Y\) into the integral, we have:

\(\(P(Z \leq z) = \int_{0}^{1} \int_{0}^{zx} 1 \cdot 1 \, dy \, dx\)\)

\(\(P(Z \leq z) = \int_{0}^{1} zx \, dx\)\(P(Z \leq z) = \left[\frac{1}{2}z^2x^2\right]_{0}^{1}\)\(P(Z \leq z) = \frac{1}{2}z^2\)\)

Now, to obtain the PDF of (Z), we differentiate the CDF with respect to (z):

\(\(f_Z(z) = \frac{d}{dz} \left(\frac{1}{2}z^2\right)\)\(f_Z(z) = z\)\)

Therefore, the PDF of Z is given by:

\(\(f_Z(z) = \begin{cases} z, & 0 \leq z \leq 1 \\ 0, & \text{otherwise} \end{cases}\)\)

This is the PDF of the random variable\(\(Z = \frac{Y}{X}\)\), where X and Y are independent random variables uniformly distributed on the interval [0,1].

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Answer:

the answer is ____

Step-by-step explanation:

first study more

then clean that computer

and then dont ask on here okau

The region enclosed by the curves y = x, y = 3x, and y = -x + 4 needs to be sketched, and its area should be found.

To sketch the region enclosed by the curves, we need to plot the three given curves on a coordinate plane. The first curve is y = x, which represents a straight line passing through the origin (0,0) with a slope of 1. The second curve is y = 3x, which is also a straight line passing through the origin but with a steeper slope of 3. The third curve is y = -x + 4, which represents a line with a y-intercept of 4 and a negative slope of -1. By plotting these three lines on the same coordinate plane, we can see that they intersect at three points: (0,0), (1,3), and (3,1). The region enclosed by these curves is a triangular region with vertices at these three points. To find the area of this triangular region, we can use the formula for the area of a triangle: A = (1/2) * base * height. Let's draw the graph:

     |

   4 |         .  (2, 2)

     |      .

     |   .

     | .

   0 |_____________________

     0     1     2     3     4     5     6

In this graph, the first equation (y = x) is depicted by a diagonal line passing through the origin (0,0). The second equation (y = 3x) is a steeper line, while the third equation (y = -x + 4) is a downward-sloping line with a y-intercept of 4. In this case, the base of the triangle is the distance between the points (0,0) and (3,1), which is 3 units. The height of the triangle is the distance between the point (1,3) and the line y = -x + 4, which is also 3 units. Substituting these values into the area formula, we get A = (1/2) * 3 * 3 = 4.5 square units.

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To find the density of Y, we need to apply the transformation method. Let's denote the random variable Y as Y = F(X).

Start by finding the cumulative distribution function (CDF) of Y, denoted as G(y):
  G(y) = P(Y ≤ y) = P(F(X) ≤ y)

To find G(y), we substitute F(X) with y and solve for X:
  F(X) ≤ y
  X ≤ F^(-1)(y)

Now, we calculate the density of Y, denoted as g(y), by differentiating G(y) with respect to y:
  g(y) = dG(y)/dy

Substitute the expression for X we found earlier into g(y):
  g(y) = d/dy [P(X ≤ F^(-1)(y))]
       = d/dy [F(F^(-1)(y))]
       = d/dy [y]
       = 1

The density of Y is 1, which means Y is a uniform random variable on the interval [0, 1].

The density of Y is 1, indicating that Y is a uniform random variable on the interval [0, 1]. The given problem asks us to find the density of Y, where Y is a continuous random variable defined as Y = F(X), and X is another continuous random variable with a strictly increasing cumulative distribution function (CDF), denoted as F(x). To solve this problem, we can use the transformation method. We start by finding the cumulative distribution function (CDF) of Y, denoted as G(y). We have G(y) = P(Y ≤ y) = P(F(X) ≤ y). To find G(y), we substitute F(X) with y and solve for X. This leads us to X ≤ F^(-1)(y). Next, we calculate the density of Y, denoted as g(y), by differentiating G(y) with respect to y. This gives us g(y) = 1. Therefore, the density of Y is 1, indicating that Y is a uniform random variable on the interval [0, 1].

The density of Y, defined as Y = F(X), is 1, which implies that Y is a uniform random variable on the interval [0, 1].

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The equation of a circle with center (a, b) and radius r is:

(x - a)^2 + (y - b)^2 = r^2

(x, y) represents any point on the circle

A circle is described as a shape consisting of all points in a plane that are at a given distance from a given point, the center.

The properties of the circle are as follows:

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X = 22

Y = 15

youre  welcome! Brainliest would be appreciated :)

The value of p is 2 and q is either 7 or 13 and their sum is either 9 or 15.

It is given that the value of p and q is a prime number, and p<q.

But it is also given that the value of both should be less than 18 as well as a prime number.

Therefore the value of both should be 2, 3, 5, 7, 11, 13, and 17.

And it is also given that p<q.

Now we have to find the value of p + q which is odd but not a prime number.

To have a sum as an odd number one number should be even and as it is given that p<q, therefore p=2, and q = 3, 5, 7, 11, 13, 17.

So

p + q = 2 + 5 = 7, which is odd but it is prime

Now again, q = 7

p + q = 2 + 7 = 9 , which is odd as well as not prime.

Again take q = 11

p + q = 2 + 11 = 13, which is odd but prime

Now take q = 13

p + q =  2 + 13 = 15, which is odd as well as not prime.

Now take q = 17

p + q = 2 + 17 = 19, which is odd and prime.

Therefore we have p =2 and q = 7 or 13 and sum p + q =9 or 15 which is odd but not prime.

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Answer:

51

Step-by-step explanation:

Order of Operations: BPEMDAS

Step 1: Define

z² + 3z - 3

z = 6

Step 2: Substitute and Evaluate

6² + 3(6) - 3

36 + 18 - 3

36 + 15

51

Answer:

Total Number of students taking Math and History and not music are 68.

Number of students who took both math and history at the same time is 6.

Step-by-step explanation:

History = 30

Math = 32

Music= 33

Math and Music= 9

History and Music= 13

History and Music and Math = 2

History but not math = 14

We have to find students taking history and math but not music.

Total number of students = 125

From the figure  number of students taking Math and History and not music are

=30+6+32= 68

Number of students who took both math and history at the same time is 6.

Answer:

Step 1: Simplify both sides of the equation.

−7+3(8n−8)=8(7+3n)

−7+(3)(8n)+(3)(−8)=(8)(7)+(8)(3n)(Distribute)

−7+24n+−24=56+24n

(24n)+(−7+−24)=24n+56(Combine Like Terms)

24n+−31=24n+56

24n−31=24n+56

Step 2: Subtract 24n from both sides.

24n−31−24n=24n+56−24n

−31=56

Step 3: Add 31 to both sides.

−31+31=56+31

0=87

Answer:

There are no solutions.

Step-by-step explanation:

Answer:

angle TVU

Step-by-step explanation:

angle UVT

please mark as brainliest

Answer:

22 is th area

Step-by-step explanation:

The  answer to the multiple choice question is that a t distribution has slightly broader tails than the Z distribution. This is because the t distribution is based on smaller sample sizes and therefore has more variability, leading to wider tails.

Additionally, the explanation for the other options is as follows:
- A t distribution can be positively or negatively skewed depending on the sample size and distribution of the data.
- A t distribution does have a nonzero mean, just like any other distribution.
- As mentioned earlier, a t distribution has slightly broader tails than the Z distribution.
- A t distribution does have asymptotic tails, meaning they approach but never reach zero as the degrees of freedom increase.

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we first look at the denominators, hmmm we have 2 and 4, and then do a quick prime factoring.

2 = 2 * 1

4 = 2 * 2 * 1

well, let's notice that 4 contains 2 already, plus another 2 and 1 so, we can just use that our LCD.

\(\cfrac{1}{2}+\cfrac{1}{4}\implies \cfrac{(2)1~~ + ~~(1)1}{\underset{\textit{using this LCD}}{4}}\implies \cfrac{2+1}{4}\implies \cfrac{3}{4}\)

Conceptually, the arithmetic and geometric average returns are different measures used to describe the performance of an investment or an asset over a specific period.

The arithmetic average return, also known as the mean return, is calculated by adding up all the individual returns and dividing by the number of periods. It represents the average return for each period independently.

On the other hand, the geometric average return, also called the compound annual growth rate (CAGR), considers the compounding effect of returns over time. It is calculated by taking the nth root of the total cumulative return, where n is the number of periods.

When to use each measure depends on the context and purpose of the analysis:

1. Arithmetic Average Return: This measure is typically used when you want to evaluate the average return for each individual period in isolation. It is useful for analyzing short-term returns, such as monthly or quarterly returns. The arithmetic average return provides a simple and straightforward way to assess the periodic performance of an investment.

2. Geometric Average Return: This measure is more suitable when you want to understand the compounded growth of an investment over an extended period. It is commonly used for long-term investment horizons, such as annual returns over multiple years.

The geometric average return provides a more accurate representation of the overall growth rate, accounting for the compounding effect and reinvestment of returns.

In summary, the arithmetic average return is suitable for analyzing short-term performance, while the geometric average return is preferred  evaluating long-term growth and the compounding effect of returns.

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Answer:

5.3, \(4\frac{1}{4}\), 3.5, \(2\frac{4}{5}\)

Step-by-step explanation:

Answer:

Step-by-step explanation:

We are going to use the areas given to find the lengths of the sides of each of their respective squares. For the purple square, the area is 35 units squared. Because the formula for the area is A = s * s, then we can fill in the value for the area and solve for s, the side length, of the purple square.

\(35=s^2\) and

\(s=\sqrt{35}\). That side length also serves as the height of the right triangle. Now on to the blue square on the bottom. It has an area of 50 units squared, so

\(50=s^2\) and

\(s=\sqrt{50}\). We could feasibly simplify that, but it's not necessary, really. That side serves as the base of the right triangle. Now we can use Pythagorean's Theorem to find the length of the hypotenuse of the right triangle, which also serves as the side of the big blue square. We will call the side of the big blue square x.

\(x^2=(\sqrt{35})^2+(\sqrt{50})^2\) and

\(x^2=35+50\) and

\(x^2=85\) and

\(x=\sqrt{85}\). That is the side length of the large blue square. The area for the square is s * s, and since we know the side length to be √85:

\(A=(\sqrt{85})(\sqrt{85})\) so

A = 85 units squared

The probability P(22 ≤ X ≤ 33) for the given binomial distribution is approximately 0.0667, using the normal approximation to the binomial distribution.

To find the probability P(22 ≤ X ≤ 33) for a binomial distribution with parameters n = 100 and p = 0.36, we need to approximate it using the normal distribution.

In this case, we can use the normal approximation to the binomial distribution, which states that for large values of n and moderate values of p, the binomial distribution can be approximated by a normal distribution with mean μ = np and standard deviation σ = √(np(1-p)).

For X ~ B(100, 0.36), the mean μ = 100 * 0.36 = 36 and the standard deviation σ = √(100 * 0.36 * (1 - 0.36)) ≈ 5.829.

To find P(22 ≤ X ≤ 33), we convert these values to standard units using the formula z = (x - μ) / σ. Substituting the values, we have z1 = (22 - 36) / 5.829 ≈ -2.395 and z2 = (33 - 36) / 5.829 ≈ -0.515.

Using the standard normal distribution table or a calculator, we can find the corresponding probabilities for these z-values. P(-2.395 ≤ Z ≤ -0.515) is approximately 0.0667.

Therefore, the probability P(22 ≤ X ≤ 33) for the given binomial distribution is approximately 0.0667.

Note that the normal approximation to the binomial distribution is valid when np ≥ 5 and n(1-p) ≥ 5. In this case, 100 * 0.36 = 36 and 100 * (1-0.36) = 64, both of which are greater than or equal to 5, satisfying the approximation conditions.

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Answer:

118 r0

Step-by-step explanation:

590/5=118

there is no remainder

hope this helps :3

if it did pls mark brainliest

The rate of change of y with respect to x when t = 3 is = 3/25 and  the answer is not one of the options given.

An object moves in the xy-plane so that its position at any time tis given by the parametric equations x(t) = t° - 3t + 2 and y (t) = /t² + 16.

To find the rate of change of y with respect to x, we need to find dy/dx, which can be computed using the chain rule of differentiation:

(dy/dt) / (dx/dt) = dy/dx

We first compute dx/dt and dy/dt:

dx/dt = 1 - 3 = -2

dy/dt = 2t / (t² + 16)

When t = 3, we have:

dx/dt = -2

dy/dt = 2(3) / (3² + 16) = 6/25

Therefore, the rate of change of y with respect to x when t = 3 is:

(dy/dt) / (dx/dt) = (6/25) / (-2) = -3/25

So, the answer is not one of the options given.

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According to the information in the graph, the money that was collected was $79.5.

To make a graph of this problem, we must relate the number of units sold of special dinners and lunches and the total price obtained for these sales.

This graph helps us because it allows us to visualize how much money we got from the sale of these.

According to the graph the total price of lunches and dinners would be $79.5.

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Answer:

64 times larger

Step-by-step explanation:

8^3=512

2^3=8

512/8=64

Answer:

A. Future Value of Jamie Lee's deposits: $49,075.20

B. Future Value of Jamie Lee's emergency fund: $3,484.40

C. Minimum amount Jamie Lee would need now: $8,001

D. Value of five years of salary: $522,576

Step-by-step explanation:

To calculate the future value of Jamie Lee's deposits, we need to use the future value of an annuity formula. The formula is:

Future Value = Regular deposit amount x Future value of annuity factor

Given that Jamie Lee is depositing $1,800 per year and the time period is 5 years with an interest rate of 2%, we can find the future value of these deposits using the table below:

Period 2% Future Value of Annuity

1 5.102

2 10.305

3 15.717

4 21.362

5 27.264

A. Future Value of a Series of Deposits:

Future Value = $1,800 x 27.264 = $49,075.20

Therefore, the future value of Jamie Lee's deposits will be $49,075.20.

B. To calculate the future value of her emergency fund, we need to use the future value of a single amount formula:

Future Value = Current amount x Future value factor

Given that her emergency fund is $3,100 and the time period is 5 years with an interest rate of 2%, we can find the future value using the table below:

Period 2% Future Value Factor

1 1.104

2 1.109

3 1.114

4 1.119

5 1.124

B. Future Value of a Single Amount:

Future Value = $3,100 x 1.124 = $3,484.40

Therefore, her emergency fund will be worth $3,484.40.

C. To calculate the minimum amount Jamie Lee would need now to ensure she has $9,000 in the future, we need to use the present value of a single amount formula:

Present Value = Future amount desired x Present value factor

Given that she wants $9,000 in the future and the time period is 5 years with an interest rate of 2%, we can find the present value using the table below:

Period 2% Present Value Factor

1 0.898

2 0.896

3 0.893

4 0.891

5 0.889

C. Present Value of a Single Amount:

Present Value = $9,000 x 0.889 = $8,001

Therefore, the minimum amount Jamie Lee would need now is $8,001 to ensure she has $9,000 in the future.

D. To calculate the value of five years of salary, we need to use the present value of a series of deposits formula:

Present Value = Regular amount to be withdrawn x Present value of annuity factor

Given that she needs $24,000 per year for 5 years and the interest rate is 2%, we can find the present value using the table below:

Period 2% Present Value of Annuity

1 4.451

2 8.818

3 13.165

4 17.486

5 21.774

D. Present Value of a Series of Deposits:

Present Value = $24,000 x 21.774 = $522,576

Therefore, the value of five years of salary when the cupcake café opens is $522,576.

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Answer:

3 is 15 degrees and 1 is 67 degrees

Answer:

17 is the value of x...

Step-by-step explanation:

9x + 7 = 10x-10 reason ( being alternate angle

or, x = 17