A problem states
there are 5 more girls than boys and girls there are 27 students in the class in all How many boys are in the class
What are the unknowns in this problem

A Number of students in the class
B The number of boys in the class
C The number of girls in class
D The number of boys and the number of girls in class

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:

yes, since 483 is greater than 121, the correct answer is a whole number with a fraction:

483/121 = 3.991, or about 4.  "483 is approx. four times 121."

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

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."

The union of sets a, b, and d (a ∪ b ∪ d) gives you the set {2, 9, 13, 28, 40}. Set 'a' is simply the set containing elements 2 and 9.

To resolve the question, we need to analyze what each symbol means. The ∪ symbol in set theory represents union, meaning everything that is in either of the sets or in both. However, it seems there is a typographical error in your question with 'bc'. As 'c' is not defined, we will proceed by ignoring that particular part and focus on 'a' which is defined.

So, if we're looking to identify a = {2,9}, it simply means the set that contains two elements: 2 and 9.

As your question stands, based on the provided sets, s = a ∪ b ∪ d = {2, 9, 9, 13, 28, 40} but when we simplify the set (since a set does not contain duplicate values), we get s = {2, 9, 13, 28, 40}.

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Answer:

The correct option is 3.

Step-by-step explanation:

Form the given figure it is noticed that the line is passing through the points (60,4) and (120,8).

[tex]y\propto x[/tex]

[tex]y=kx[/tex]

Where, k is the constant of proportionality or slope.

The slope of a line is defined as

[tex]k=\frac{y_2-y_1}{x_2-x_1}[/tex]

[tex]k=\frac{8-4}{120-60}[/tex]

[tex]k=\frac{4}{60}[/tex]

[tex]k=\frac{1}{15}[/tex]

Therefore option 3 is correct.

Answer:

D

Step-by-step explanation:

The missing values are 12 and 18.

The ratios are  [tex]\dfrac{9}{12}[/tex] and  [tex]\dfrac{18}{24}[/tex].

The given table is:

[tex]\begin{center}\begin{tabular}{ c c c c } shoes & 36 & 9 & y \\ socks & 48 & x & 24 \\\end{tabular}\end{center}[/tex]

Since all the columns are pertaining same ratio; thus we have:

[tex]\dfrac{36}{24} = \dfrac{9}{x} = \dfrac{y}{24}\\\\\dfrac{3}{4} = \dfrac{9}{x} = \dfrac{y}{24}\\\\\\or\\\\\dfrac{3}{4} = \dfrac{9}{x}\\\\x = 12\\\\and \\\\\dfrac{3}{4} = \dfrac{y}{24}\\\\y = 18[/tex]

Thus, the missing values x and y are 12 and 18 respectively.

And the are  ratios  [tex]\dfrac{9}{12}[/tex] and  [tex]\dfrac{18}{24}[/tex].

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We are given that there are 5 groups which are:

group of 2

group of 3

group of 4

group of 5

group of 6

and 1 left over

So the smallest number of cupcakes would simply be the sum of all:

smallest number of cupcakes = 2 + 3 + 4 + 5 + 6 + 1

smallest number of cupcakes = 21

Answer:

They raised $150

Step-by-step explanation:

In order to solve this you just need to reverse the actions that they took in order to get to 47 dollars left, so the total amount they raised will be represented by letter X, so 47 is what is left after spending 29 on trash bags and 74 on waterproof boots, that means that from the total we are withdrawing those amounts and the result will be 47:

Total-Money spent=47

X-29 - 74= 47

x=47+74+29

x=150

So now we know that they originally had 150 dollars and that would be what they raised.

Final answer:

By using the compound interest formula A=P(1+r)² and the given values, we find the interest rate to be approximately 0.09, or 9% annually.

Explanation:

To find the interest rate r that grew the principal P from $5000 to $5940.50 over two years with compound interest, we use the formula A=P(1+r)². Here, A is the amount of money accumulated after n years, including interest. We are given that A is $5940.50 and P is $5000.

Let's plug in the values and solve for r:

5940.50 = 5000(1+r)²
1.1881 = (1+r)²

To find r, we take the square root of 1.1881:

Subtracting 1 from both sides to isolate r, we get:

r = 1.09 - 1
r = 0.09

The approximate interest rate is 0.09, which means the annual interest rate is 9%.

the correct answer is A. [tex]\( f(x) = 2x + 1 \).[/tex]

To find the rule for the linear function represented in the table, we can use the formula for a linear function, which is:

[tex]\[ f(x) = mx + b \][/tex]

Where:

- m is the slope of the line

- b is the y-intercept

Given the table:

[tex]\[ \x & : 2, 5, 8, 11 \\f(x) & : 5, 11, 17, 23\end{align}\][/tex]

We can start by finding the slope (m) using the formula:

[tex]\[ m = \frac{{f(x_2) - f(x_1)}}{{x_2 - x_1}} \][/tex]

Let's choose two points from the table, for example, (2, 5) and (5, 11):

[tex]\[ m = \frac{{11 - 5}}{{5 - 2}} \]\[ m = \frac{{6}}{{3}} \]\[ m = 2 \][/tex]

So, we have found that the slope m is 2.

Now, we can use the slope-intercept form of a line to find the y-intercept (b). We can pick any point from the table to do this. Let's use the point (2, 5):

f(x) = mx + b

5 = 2(2) + b

5 = 4 + b

b = 5 - 4

b = 1

So, we have found that the y-intercept b is 1.

Now, we can write the rule for the linear function:

[tex]\[ f(x) = 2x + 1 \][/tex]

Therefore, the correct answer is A. [tex]\( f(x) = 2x + 1 \).[/tex]

The complete question is:

Write a rule for the linear function in the table.

x = 2,5,8,11

f(x) = 5,11,17,23

A. f(x) = 2x + 1

B. f(x) = x + 5

C. f(x) = –2x – 1

D. f(x) = 1/2x+1