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Sergio039 [100]

2 years ago

6

For two days in a row Alvin jogged in the park he jogged for 46 min on Friday this is 19 min less than he ran on Saturday write

and solve a subtraction equation to find the amount of time Alvin spent jogging on Saturday

1 answer:

Luda [366]2 years ago

8 0

Saturday = 46+19 min = 65 minutes

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Mrs.Steﬀen’s third grade class has 30 students in it. The students are divided into three groups(numbered 1, 2,and 3),each havin

qaws [65]

Answer:

a. $\\ 10! = 3628800$ ;

b. $\\ 10!*10!*10! = 47784725839872000000 = 4.7784725839872*10^{19}$

Step-by-step explanation:

We need here to apply the <em>Multiplication Principle </em>or the <em>Fundamental Principle of Counting</em> for each answer. Answer <em>b</em> needs an extra reasoning for being completed.

The <em>Multiplication Principle</em> states that if there are <em>n</em> ways of doing something and <em>m</em> ways of doing another thing, then there are <em>n</em> x <em>m</em> ways of doing both (<em>Rule of product</em> (2020), in Wikipedia).

<h3>In how many ways can ten students line up? </h3>

There are <em>ten</em> students. When one is selected, there is no other way to select it again. So, <em>no repetition</em> is allowed.

Then, in the beginning, there are 10 possibilities for 10 students; when one is selected, there are nine possibilities left. When another is selected, eight possibilities are left to form the file, and so on.

Thus, we need to multiply the possibilities after each selection: that is <em>why</em> the <em>Multiplication Principle</em> is important here.

This could be expressed mathematically using n!:

$\\ n! = n * (n-1)! * (n-2)! *...* 2*1.$

For instance, $\\ 5! = 5 * (5-1)! * (5-2)! *...*2*1 = 5 * 4 * 3 * 2 * 1 = 120.$

So, for the case in question, the <em>ten</em> students can line up in:

$\\ 10! = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 3628800$ ways to line up in a single file.

<h3>Second Question</h3>

For this question, we need to consider the former reasoning with extra consideration in mind.

The members of Group 1 can occupy <em>only</em> the following places in forming the file:

$\\ G1 = \{ 1, 4, 7, 10, 13, 16, 19, 22, 25, 28\}^{th}$ <em>places</em>.

The members of Group 2 <em>only</em>:

$\\ G2 = \{ 2, 5, 8, 11, 14, 17, 20, 23, 26, 29\}^{th}$ <em>places</em>.

And the members of Group 3, the following <em>only</em> ones:

$\\ G3 = \{ 3, 6, 9, 12, 15, 18, 21, 24, 27, 30\}^{th}$ <em>places.</em>

Well, having into account these possible places for each member of G1, G2 and G3, there are: <em>10! ways</em> for lining up members of G1; <em>10! ways</em> for lining up members of G2 and, also, <em>10! ways</em> for lining up members of G3.

After using the <em>Multiplication Principle</em>, we have, thus:

$\\ 10! * 10! * 10! = 47784725839872000000 = 4.7784725839872 *10^{19}$ <em>ways the students can line up to come in from recess</em>.

3 0

1 year ago

. Cara is playing a game. For each dart she can toss onto a board, she

Kaylis [27]

She will need 12 seconds because -20•12=-240

6 0

2 years ago

Ray Np is an angle bisector of angle MNQ and measure of Angle PNQ = 2x+1. Find the measure of angle MNQ. If The measure of MNQ=x

Step2247 [10]

Refer to the diagram below.

Because ray NP bisects ∠MNQ, therefore
∠MNP = ∠PNQ = 2x + 1.
Therefore
∠MNQ = 2*∠PNQ = 2(2x + 1) = 4x + 2.

Because ∠MNQ is given as x² - 10, therefore
x² - 10 = 4x + 2
x² - 4x - 12 = 0
(x + 2 )(x - 6) = 0
x = -2, or x = 6

When x = -2,
∠MNQ = 4*(-2) + 2 = -6°
This answer is not acceptablle, therefore x = -2 should be rejected.

When x = 6,
∠MNQ = 4*6 + 2 = 26°

Answer: x = 6, and ∠MNQ = 26°

7 0

2 years ago

A coordinate plane showing Running. The x-axis shows seconds, the y-axis shows feet. One solid line showing Moises starting at (

torisob [31]

Answer:

Step-by-step explanation:

s

7 0

1 year ago

(1 point) Let P(t) be the performance level of someone learning a skill as a function of the training time t. The derivative dPd

monitta

Answer:

$P(t)=M+Ce^{-kt}$

Step-by-step explanation:

Given the differential model

$\dfrac{dP}{dt}=k[M-P(t)]$

We are required to solve the equation for P(t).

$\dfrac{dP}{dt}=kM-kP(t)\\$Add kP(t) to both sides\\\dfrac{dP}{dt}+kP(t)=kM\\$Taking the integrating factor\\e^{\int k dt} =e^{kt}\\$Multiply all through by the integrating factor\\\dfrac{dP}{dt}e^{kt}+kP(t)e^{kt}=kMe^{kt}\\\dfrac{dP}{dt}e^{kt}=kMe^{kt}\\(Pe^{kt})'=kMe^{kt} dt\\$Take the integral of both sides with respect to t\\\int (Pe^{kt})'=\int kMe^{kt} dt\\Pe^{kt}=kM \int e^{kt} dt\\Pe^{kt}=\dfrac{kM}{k} e^{kt} + C_0, C_0$ a constant of integration$

$Pe^{kt}=Me^{kt} + C\\$Divide both side by e^{kt}\\P(t)=M+Ce^{-kt}\\P(t)=M+Ce^{-kt}\\$Therefore:\\P(t)=M+Ce^{-kt}$

4 0

2 years ago