All categories

2 years ago

Distance run by each member

Mathematics

1 answer:

saw5 [17]2 years ago

8 0

Answer:

The median is the middle term in the arranged data set. ... The median class interval is the corresponding class where the median value falls.

Step-by-step explanation:

The median is the middle term in the arranged data set. ... The median class interval is the corresponding class where the median value falls.

You might be interested in

Estimate the quotient what is 15.1 divided by 7

Zolol [24]

15.1 is close to 14 so then you have 14/7 which approximately = 2

8 0

2 years ago

Read 2 more answers

A sports store near Big Bear Lake is having a 20% off sale on all water skis. What will the sale price be for water skis which r

kvasek [131]

248 x 0.8 = $198.40
answer $198.40

8 0

2 years ago

The local radio tower is located 20 miles east and 12 miles north of Pamela's cabin. The radio tower has a 25 mile broadcast rad

pychu [463]

Answer:

x^2 +y^2 -40x -24y -81 =0

Step-by-step explanation:

We can write the equation of a circle in the form

(x-h)^2 + (y-k)^2 = r^2

Where (h,k) is the center and r is the radius

Assuming that Pamela's cabin is the center

( x-20)^2 + (y- 12)^2 = 25^2

We want to get to general form

x^2 + y^2 + Dx + Ey + F = 0,

Distributing

x^2 -40x +400 + y^2 -24y +144 = 625

x^2 +y^2 -40x -24y +544 -625 =0

x^2 +y^2 -40x -24y -81 =0

3 0

2 years ago

Write down the explicit solution for each of the following: a) x’=t–sin(t); x(0)=1

Kay [80]

Answer:

a) x=(t^2)/2+cos(t), b) x=2+3e^(-2t), c) x=(1/2)sin(2t)

Step-by-step explanation:

Let's solve by separating variables:

$x'=\frac{dx}{dt}$

a) x’=t–sin(t), x(0)=1

$dx=(t-sint)dt$

Apply integral both sides:

$\int {} \, dx=\int {(t-sint)} \, dt\\\\x=\frac{t^2}{2}+cost +k$

where k is a constant due to integration. With x(0)=1, substitute:

$1=0+cos0+k\\\\1=1+k\\k=0$

Finally:

$x=\frac{t^2}{2} +cos(t)$

b) x’+2x=4; x(0)=5

$dx=(4-2x)dt\\\\\frac{dx}{4-2x}=dt \\\\\int {\frac{dx}{4-2x}}= \int {dt}\\$

Completing the integral:

$-\frac{1}{2} \int{\frac{(-2)dx}{4-2x}}= \int {dt}$

Solving the operator:

$-\frac{1}{2}ln(4-2x)=t+k$

Using algebra, it becomes explicit:

$x=2+ke^{-2t}$

With x(0)=5, substitute:

$5=2+ke^{-2(0)}=2+k(1)\\\\k=3$

Finally:

$x=2+3e^{-2t}$

c) x’’+4x=0; x(0)=0; x’(0)=1

Let $x=e^{mt}$ be the solution for the equation, then:

$x'=me^{mt}\\x''=m^{2}e^{mt}$

Substituting these equations in c)

$m^{2}e^{mt}+4(e^{mt})=0\\\\m^{2}+4=0\\\\m^{2}=-4\\\\m=2i$

This becomes the solution m=α±βi where α=0 and β=2

$x=e^{\alpha t}[Asin\beta t+Bcos\beta t]\\\\x=e^{0}[Asin((2)t)+Bcos((2)t)]\\\\x=Asin((2)t)+Bcos((2)t)$

Where A and B are constants. With x(0)=0; x’(0)=1:

$x=Asin(2t)+Bcos(2t)\\\\x'=2Acos(2t)-2Bsin(2t)\\\\0=Asin(2(0))+Bcos(2(0))\\\\0=0+B(1)\\\\B=0\\\\1=2Acos(2(0))\\\\1=2A\\\\A=\frac{1}{2}$

Finally:

$x=\frac{1}{2} sin(2t)$

7 0

2 years ago

7. (05.02 MC)

Diano4ka-milaya [45]

Considering the Central Limit Theorem, we have that:

a) The probability cannot be calculated, as the underlying distribution is not normal and the sample size is less than 30.

b) The probability can be calculated, as the sample size is greater than 30.

<h3>What does the Central Limit Theorem state?</h3>

It states that the sampling distribution of sample means of size n is approximately normal has standard deviation $s = \frac{\sigma}{\sqrt{n}}$ , as long as the underlying distribution is normal or the sample size is greater than 30.

In this problem, the underlying distribution is skewed right, that is, not normal, hence:

For item a, the probability cannot be calculated, as the underlying distribution is not normal and the sample size is less than 30.

For item b, the probability can be calculated, as the sample size is greater than 30.

More can be learned about the Central Limit Theorem at brainly.com/question/16695444

#SPJ1

8 0

1 year ago