Ten pairs of points yielded a correlation coefficient r of 0.790. If a =0.05, which of the following statements is correct if H.: P = 0? (Do not calculate a t-value.) A) Because 0.790 is greater than 0.632, the nullliy pothesis is not rejected. Because 0.790 is greater than 0.602, che null hypothesis is not rejected. Because 0.790 is greater than 0.632, che null hypothesis is rejected. OD) There is no correlation between the variables​

Answer:

See Below.

Step-by-step explanation:

We are given that:

[tex]\displaystyle \frac{dT}{dt} = -k(T - T_0)[/tex]

And we want to show that:

[tex]\displaystyle T = T_0+Ae^{-kt}[/tex]

From the original equation, divide both sides by (T - T₀) and multiply both sides by dt. Hence:

[tex]\displaystyle \frac{dT}{T-T_0}= -k\, dt[/tex]

Take the integral of both sides:

[tex]\displaystyle \int \frac{dT}{T- T_0} = \int -k \, dt[/tex]

Integrate. For the left integral, we can use u-substitution. Note that T₀ is simply a constant. Hence:

[tex]\displaystyle \ln\left|T - T_0\right| = -kt+C[/tex]

Raise both sides to e:

[tex]\displaystyle e^{\ln\left|T-T_0\right|} = e^{-kt+C}[/tex]

Simplify:

[tex]\displaystyle \begin{aligned} \left| T- T_0\right| &= e^{-kt} \cdot e^C \\ \\ &= e^C\left(e^{-kt}\right) \\ \\ &=Ae^{-kt} & \text{Let $e^C = A$}\end{aligned}[/tex]

Since the temperature T will always be greater than or equal to the surrounding medium T₀, we can remove the absolute value. Hence:

[tex]\left(T - T_0\right) = Ae^{-kt}[/tex]

Therefore:

[tex]\displaystyle T = T_0+Ae^{-kt}[/tex]

Mandatory minimum character count of 20.

Answer:

You can cut the plank down from the size you buy to the

exact size, but you want to waste as little wood as possible.

Step-by-step explanation:

D. RAMONA SAVED THE MOST IN 2006

D. Ramona saved the most in 2006

Answer:

Step-by-step explanation:

its easyk

[tex]\displaystyle \int\sec x\:dx = \ln |\sec x + \tan x| + C[/tex]

Step-by-step explanation:

[tex]\displaystyle \int\sec x\:dx=\int\sec x\left(\frac{\sec x+ \tan x}{\sec x + \tan x}\right)dx[/tex]

[tex]\displaystyle = \int \left(\dfrac{\sec x\tan x + \sec^2x}{\sec x + \tan x} \right)dx[/tex]

Let [tex]u = \sec x + \tan x[/tex]

[tex]\:\:\:\:\:\:du = (\sec x\tan x + \sec^2x)dx[/tex]

where

[tex]d(\sec x) = \sec x\tan x\:dx[/tex]

[tex]d(\tan x) = \sec^2x\:dx[/tex]

[tex]\displaystyle \Rightarrow \int \left(\frac{\sec x\tan x + \sec^2x}{\sec x + \tan x}\right)\:dx = \int \dfrac{du}{u}[/tex]

[tex]= \ln |u| + C = \ln |\sec x + \tan x| + C[/tex]

Answer:

141

Step-by-step explanation:

if the sum of the two angles equals 180 subtract 39 from 180 to get the remainder of 141 which is angle 2

Answer:

0.02 m/sec

Step-by-step explanation:

26/30=0.89 —> 0.89 min —> 53.4 sec

42/50=0.84 meters

speed=0.84 / 53.4 = 0.015 m/sec = 0.02 m/sec

Answer:

x is 1. i looked it up so that's all you need

Answer:

Step-by-step explanation:

Substituting y = 4x+1 into 2y = -x-1 gives the equation

2(4x+1) = -x-1

Solve the equation:

8x+2 = -x-1

9x = -3

x = -⅓

Substituting y = 4x+1 into 2y = -x-1 will produce the equation 2(4x+1) = -x-1

A mathematical statement known as an equation is made up of two expressions joined together by the equal sign. Based on the degree, there are four different main types of equations. Equations that are linear, quadratic, cubic, and polynomial

Given,  the equation y = 4x + 1 and another equation 2y = -x – 1.

Substituting equation 1 into equation 2 we will get

2(4x+1) = -x-1

Solve the equation:

8x+2 = -x-1

9x = -3

x = -⅓

Therefore, The equation 2(4x+1) = -x-1 is created when y = 4x+1 is substituted into 2y = -x-1.

Learn more about equations here:

https://brainly.com/question/16255566

#SPJ2

Answer:

96 employees

Step-by-step explanation:

Given that the standard deviation = 22.8

The width in the question = 12

We solve for the margin of error E.

E = width / 2

= 12/2 = 6

At 99%

Alpha = 1-0.99

= 0.01

Alpha/2 = 0.01/2 = 0.005

Z0.005 = 2.576

Sample size n

= ((2.576x22.8)/2)²

= 95.8

= 96

The number of employees is 96

Thank you!

Answer:

[tex]f(-1) = -2[/tex]

Step-by-step explanation:

Given

The attached graph

Required

[tex]f(-1)[/tex]

This is the point where

[tex]x = -1[/tex]

On the attached graph;

[tex]f(x) = -2[/tex] when [tex]x = -1[/tex]

Hence:

[tex]f(-1) = -2[/tex]

Answer:

2.7

Step-by-step explanation:

ratios help

2.5 : 5.8 :: x : 6.2

2.5/5.8 = x/6.2

solve for x :

x = approx.   2.7

Answer:

0.1492-0.1515= -0.0023

Answer:

-a^2 +ab +3a

Step-by-step explanation:

-a(a-b-3)

Distribute

-a*a -a*(-b) -a *(-3)

-a^2 +ab +3a

This is the answers and their coordinates

9514 1404 393

Answer:

  C.  159°

Step-by-step explanation:

The exterior angle at B is half the difference of the measures of the arcs it intercepts:

  (3x +19)° = 1/2((17x -3)° -91°)

  6x +38 = 17x -94 . . . . . . . . . . multiply by 2, divide by °

  132 = 11x . . . . . . . . . . . . . add 94-6x

  x = 12 . . . . . . . . . . . . divide by 11

Then long arc AD is ...

  arc AD = (17(12) -3)° = 201°

Arc DCA is the rest of the circle:

  arc DCA = 360° -201° = 159°

Step-by-step explanation:

C + ſa6+b5 bescribe the Mathematical order of Operation

Answer:

{(2,0),(-5,1),(7,8),(6,0),(-7,1)

Answer:

A.

Step-by-step explanation:

x-2y=4 has a x-intercept of 4, a slope of 1/2, and a y-intercept of -2. 2x+y=4 has a x-intercept of -2, a slope of 2, and a y-intercept of -4.

9514 1404 393

Answer:

  sin(θ) = (-7√65)/65

  csc(θ) = (-√65)/7

Step-by-step explanation:

The angle will have the given characteristics if its terminal ray passes through the 3rd-quadrant point (-4, -7). The distance from the origin to that point is ...

  d = √((-4)² +(-7)²) = √65

The sine of the angle is the ratio of the y-coordinate to this value:

  sin(θ) = -7/√65

  sin(θ) = (-7√65)/65

The cosecant is the inverse of the sine

  csc(θ) = (-√65)/7

Answer:

False, true

Step-by-step explanation:

I’m not 100% sure but I think that’s the answer

Answer:

272$

Step-by-step explanation:

You really should be clearer with your questions, but if your looking for the balance she has 272$

Answer:

b) y=x -2(w+z)

Step-by-step explanation:

multiply both sides, move the terms and write on parametric form

Answer:

1.6 Bitcoins

Step-by-step explanation:

Given data

We have the rate as

$1 USD to 0.004

Hence $400 will buy x bitcoins

Cross multiply to find the value of x

1*x= 400*0.004

x=1.6

Hence $400 will get you 1.6 Bitcoins

Answer:

-(x + 3)(x - 3)

Step-by-step explanation:

Using the difference of squares we can factor this expression.

[tex](9 - x^2)\\= (3^2 - x^2)\\= (3 + x)(3 - x)\\= -(3 + x)(-3 + x)\\= -(x + 3)(x - 3)[/tex]

Answer:

[tex]f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.[/tex]

Step-by-step explanation:

Given

[tex]9 < x < 10[/tex] --- interval

Required

The probability density of the volume of the cube

The volume of a cube is:

[tex]v = x^3[/tex]

For a uniform distribution, we have:

[tex]x \to U(a,b)[/tex]

and

[tex]f(x) = \left \{ {{\frac{1}{b-a}\ a \le x \le b} \atop {0\ elsewhere}} \right.[/tex]

[tex]9 < x < 10[/tex] implies that:

[tex](a,b) = (9,10)[/tex]

So, we have:

[tex]f(x) = \left \{ {{\frac{1}{10-9}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.[/tex]

Solve

[tex]f(x) = \left \{ {{\frac{1}{1}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.[/tex]

[tex]f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.[/tex]

Recall that:

[tex]v = x^3[/tex]

Make x the subject

[tex]x = v^\frac{1}{3}[/tex]

So, the cumulative density is:

[tex]F(x) = P(x < v^\frac{1}{3})[/tex]

[tex]f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.[/tex] becomes

[tex]f(x) = \left \{ {{1\ 9 \le x \le v^\frac{1}{3} - 9} \atop {0\ elsewhere}} \right.[/tex]

The CDF is:

[tex]F(x) = \int\limits^{v^\frac{1}{3}}_9 1\ dx[/tex]

Integrate

[tex]F(x) = [v]\limits^{v^\frac{1}{3}}_9[/tex]

Expand

[tex]F(x) = v^\frac{1}{3} - 9[/tex]

The density function of the volume F(v) is:

[tex]F(v) = F'(x)[/tex]

Differentiate F(x) to give:

[tex]F(x) = v^\frac{1}{3} - 9[/tex]

[tex]F'(x) = \frac{1}{3}v^{\frac{1}{3}-1}[/tex]

[tex]F'(x) = \frac{1}{3}v^{-\frac{2}{3}}[/tex]

[tex]F(v) = \frac{1}{3}v^{-\frac{2}{3}}[/tex]

So:

[tex]f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.[/tex]