Norma and Rene are serving cupcakes at a school party. If they arrange the cupcakes in groups of 2.3.4.5. or 6 they have exactly one cupcake left over. what is the smallest number of cupcakes they could have?

The expression 6 – (–8) will be equivalent to 14.

An expression is a combination of some mathematical symbol such that arithmetic operator and variable such that all are constrained and create an equation.

In other meaning, expression is very useful to determine the end or root value of constraint.

For example 3x +5y

The constraint of expression is represented as = or <, >.

Given the expression,

6 – (–8)

By multiplication,

Negative signs become positive

6 – (–8) = 6 + 8 = 14

Hence "The expression 6 – (–8) will be equivalent to 14.".

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Answer:

As per the statement:

If a scatter plot is made with the number of chirps on the horizontal axis and

the trend line is given by:

[tex]y = 3x+25[/tex]              ....[1]

where,

y represents the temperature in degrees Fahrenheit

x represents the number of chirps per second.

We have to find  the number of chirps per second to be when the temperature is 55 degrees Fahrenheit.

Substitute y = 55 degree Fahrenheit in [1] we have;

[tex]55 = 3x+25[/tex]

Subtract 25 from both sides we have;

[tex]30= 3x[/tex]

Divide both sides by 3 we have;

10 = x

or

x = 10

Therefore, the number of chirps per second to be when the temperature is 55 degrees Fahrenheit is, 10

Answer:

-7

Step-by-step explanation:

The graphical solution to a system of two linear equations with the point (-4,5) as the only solution will show two lines intersecting at this point on a two-dimensional graph.

If the point (-4,5) is the only solution to a system of two linear equations, this means that the two lines represented by these equations intersect at this point. On a Two-Dimensional (x-y) Graphing system, this point would be plotted on the graph where the x-coordinate is -4 and the y-coordinate is 5.

These two equations when plotted as lines in a two-dimensional space (with x as the independent variable and y as the dependent variable) will intersect at this point (-4,5). If you draw a line from the origin to the point (-4,5), it will intersect the two lines at this point. This is how the graphical solution would look like.

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Answer:

45% probability that Marlene will lose her next tennis match.

Step-by-step explanation:

For each match that Marlene plays, there is a 55% probability that she wins and a 45% probability that she loses.

These probabilities for each match are independent.

So there is a 45% probability that Marlene will lose her next tennis match.

To answer this problem, we simply have to refer to the standard normal probabilities table to locate for the P value at specified z score value.

So at a value of z = 1.3, the value of P using right tailed test is:

P = 0.0968 = 9.68%

So this actually means that 9.68% of the people has a higher score

Answer:

The two positive numbers whose product is 81 and whose sum is a minimum are 9 and 9.

Step-by-step explanation:

81 is divisible by 1 and 81, since 81/1 = 81 and 81/81 = 1.

81 is divisible by 3 and 27, since 81/3 = 27 and 81/27 = 3.

81 is divisible by 9, since 81/9 = 9.

Their sums are.

81 + 1 = 82.

3 + 27 = 30

9 + 9 = 18.

So the two positive numbers whose product is 81 and whose sum is a minimum are 9 and 9.

Answer:

The two positive numbers will be 9 and 9.

Step-by-step explanation:

Let the two positive numbers be x and y.

It is given that the product of the number is 81 the sum of them is minimum.

So, the product can be written as [tex]xy=81[/tex].

The sum can be written as [tex]x+y=x+\dfrac{81}{x}[/tex]. It is required to minimize the sum.

Differentiate the sum as,

[tex]x+y=x+\dfrac{81}{x}\\1+y'=1-\dfrac{81}{x^2}=0\\x^2=81\\x=9\\y=\dfrac{81}{x}\\y=9[/tex]

Now, the second derivative of the sum will be,

[tex]y''=2\times \dfrac{81}{x^3}>0[/tex]

So, the second derivative is positive and hence, the sum will be minimum at x=9.

Therefore, the two positive numbers will be 9 and 9.

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Using the half-life and initial concentration of Atenolol, we can find the quantities p2, p3, p4,...,pn before each dose using the formula p(y+Ay)-p(y)/Ay. Without specific values, we can't provide a closed-form expression for pn.

To find the series of quantities p2, p3, p4, ..., and pn representing the amount of atenolol in the body before taking each respective dose, we would start by invoking the definition of half-life, represented as t1/2. Using half-life would mean that the concentration of A (atenolol) is one-half its initial concentration [t = t1/2, A = [4]].

The formula to find the respective concentrations would be p(y + Ay) - p(y) / Ay, where Ay is the change in amounts of Atenolol.

To find pn in closed-form, we apply the formula iteratively, starting from p2 and proceeding to pn. For example, to find p2, p3 and so forth, we'd use the previously calculated value (i.e. for calculating p3, we'd use the calculated value of p2 in the formula).

However, without specific information about the half-life of atenolol in the body and how it changes with each dose, or the exact initial concentration, we can't provide a specific expression for pn in closed-form. Generally, the expression for pn will depend on the half-life and initial concentration of Atenolol in the body.

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The expression for [tex]\( p_n \)[/tex] in closed-form is [tex]\( p_n = \frac{D}{k} (1 - e^{-nk\tau}) \)[/tex], where [tex]\( D \)[/tex] is the dose of atenolol, [tex]\( k \)[/tex] is the rate constant for elimination, and [tex]\( \tau \)[/tex] is the time interval between doses.

To derive the expression for[tex]\( p_n \)[/tex], we start by considering the pharmacokinetic model for atenolol, which can be described by the following first-order differential equation representing the rate of change of the drug concentration in the body:

[tex]\[ \frac{dp}{dt} = -kp + D\delta(t - n\tau) \][/tex]

where:

- [tex]\( p \)[/tex] is the amount of atenolol in the body at time [tex]\( t \)[/tex],

- [tex]\( k \)[/tex] is the rate constant for elimination,

-[tex]\( D \)[/tex] is the dose of atenolol administered at each time interval [tex]\( n\tau \)[/tex],

- [tex]\( \delta(t - n\tau) \)[/tex] is the Dirac delta function representing the administration of the dose at time [tex]\( n\tau \)[/tex],

-[tex]\( n \)[/tex] is the number of doses administered,

- [tex]\( \tau \)[/tex] is the time interval between doses.

For the time period right before taking the [tex]\( n \)[/tex]-th dose, we are interested in the amount of atenolol in the body at time [tex]\( t = n\tau^- \)[/tex], just before the [tex]\( n \)[/tex]-th dose is taken. We can solve the differential equation for [tex]\( p \)[/tex] during the interval[tex]\( (n-1)\tau \leq t < n\tau \)[/tex] by integrating from [tex]\( (n-1)\tau \) to \( t \)[/tex]:

[tex]\[ \int_{(n-1)\tau}^{t} \frac{dp}{dt} \, dt = -\int_{(n-1)\tau}^{t} kp \, dt \][/tex]

Since there is no input of the drug during this interval, the delta function does not contribute to the integral. Solving the integral, we get:

[tex]\[ p(t) - p((n-1)\tau) = -k \int_{(n-1)\tau}^{t} p(t) \, dt \][/tex]

Let \( p((n-1)\tau) = p_{n-1} \) be the amount of atenolol in the body right before taking the \( (n-1) \)-th dose. The solution to the above differential equation is of the form:

[tex]\[ p(t) = p_{n-1} e^{-k(t - (n-1)\tau)} \][/tex]

Now, we need to find the expression for [tex]\( p_{n-1} \).[/tex] We know that right after taking the [tex]\( (n-1) \)[/tex]-th dose, the amount of atenolol in the body is [tex]\( p_{n-1} + D \)[/tex]. As time progresses to [tex]\( t = n\tau^- \)[/tex], this amount decays to [tex]\( p_{n-1} e^{-k(n\tau - (n-1)\tau)} \)[/tex], which simplifies to [tex]\( p_{n-1} e^{-k\tau} \)[/tex].

We can now write a recursive relationship for [tex]\( p_n \)[/tex]:

[tex]\[ p_n = (p_{n-1} + D) e^{-k\tau} \][/tex]

To find the closed-form expression, we need to sum up the contributions of all previous doses, taking into account the decay factor [tex]\( e^{-k\tau} \)[/tex] for each dose:

[tex]\[ p_n = D e^{-k\tau} + D e^{-2k\tau} + \ldots + D e^{-nk\tau} \][/tex]

This is a geometric series with the common ratio [tex]\( e^{-k\tau} \)[/tex]. The sum of a geometric series is given by:

[tex]\[ S = \frac{a(1 - r^n)}{1 - r} \][/tex]

where \( a \) is the first term and [tex]\( r \)[/tex] is the common ratio. Applying this formula to our series, we get:

[tex]\[ p_n = \frac{D(1 - e^{-nk\tau})}{1 - e^{-k\tau}} \][/tex]

Multiplying the numerator and the denominator by [tex]\( e^{k\tau} \)[/tex] to simplify, we obtain:

[tex]\[ p_n = \frac{D e^{k\tau}(1 - e^{-nk\tau})}{e^{k\tau} - 1} \][/tex]

Since[tex]\( e^{k\tau} - 1 \)[/tex] is equivalent to [tex]\( k\tau \)[/tex] for small[tex]\( k\tau \)[/tex], the expression simplifies to:

 [tex]\[ p_n = \frac{D}{k} (1 - e^{-nk\tau}) \][/tex]

This is the closed-form expression for [tex]\( p_n \)[/tex], representing the amount of atenolol in the body right before taking the [tex]\( n \)-[/tex]th dose."