Is it true that Confidence intervals are always close to their true population values?

Answers

Answer 1

Answer:

Answer:

This means that there is a 95% probability that the confidence interval will contain the true population mean. Thus, P( [sample mean] - margin of error < μ < [sample mean] + margin of error) = 0.95.

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The mean breaking strength of a ceramic insulator must be at least 10 psi. The process by which this insulator is manufactured must show equivalence to this standard. If the process can manufacture insulators with a mean breaking strength of at least 9.5 psi, it will be considered equivalent to the standard. A random sample of 50 insulators is available, and the sample mean and standard deviation of breaking strength are 9.32 psi and 0.21 psi, respectively. a. State the appropriate hypotheses that must be tested to demonstrate equivalence.

Answers

Answer:

H0: u ≥ 9.5

Ha: u < 9.5

Step-by-step explanation:

The null hypothesis and the alternate hypothesis are reverse of each other.

The claim is either set as the null or alternate hypothesis

The null hypothesis is : H0: u ≥ 9.5 the mean breaking strength of a ceramic insulator is at least 9.5 which is considered equivalent to the standard.

The alternate hypothesis is: Ha: u < 9.5 the mean breaking strength of a ceramic insulator is less than 9.5 which is not considered equivalent to the standard.

Which table represents the statement “An airplane is flying at a speed of 525 miles per hour “?

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Answer: B

Step-by-step explanation:

I got it right

Answer:

The correct answer is B. d = 525h

An office building loses a third of its heat between sundown and midnight and an additional half of the original amount of heat between midnight and 4 AM. If five-eighths of the remaining heat is lost between 4 AM and 5 AM, what proportion of the total heat loss occurs between 5 AM and sunrise?

Answers

Answer:

$(1)/(16)$

Step-by-step explanation:

Proportion of Heat Loss Between sundown and midnight $=(1)/(3)$

Proportion of Heat Loss between midnight and 4 AM $=(1)/(2)$

Proportion of Total Heat Already Lost $=(1)/(3)+(1)/(2) =(5)/(6)$

Proportion of Remaining Heat $=1-(5)/(6)=(1)/(6)$

Between 4 AM and 5 AM, five-eighths of the remaining heat is lost.

Proportion of Heat Loss between 4 AM and 5 AM= $(5)/(8)$ X (1)/(6) = (5)/(48)$

Therefore, Proportion of Remaining Heat Left $=(1)/(6)- (5)/(48)=(1)/(16)$

We therefore say that:

$(1)/(16)$ of the total heat loss occurs between 5 AM and sunrise.$

Use implicit differentiation to find an equation of the tangent line to the curve at the given point. y 2 ( y 2 − 4 ) = x 2 ( x 2 − 5 ) , ( 0 , − 2 ) (devil's curve) y2(y2-4)=x2(x2-5), (0,-2) (devil's curve)

Answers

Answer:

y = -2

Step-by-step explanation:

To find the equation of the tangent we apply implicit differentiation, and then we take apart dy/dx

The equation is

$y^2(y^2-4)=x^2(x^2-5)$

implicit differentiation give us

$(d)/(dx)[y^2(y^2-4)=x^2(x^2-5)]\n\n2y(dy)/(dx)(y^2-4)+y^2(2y(dy)/(dx))=2x(x^2-5)+x^2(2x)\n\n4y^3(dy)/(dx)-8y(dy)/(dx)=2x^3-10x+2x^3\n\n(dy)/(dx)=(4x^3-10x)/(4y^3-8y)$

But we know that

$m=(dy)/(dx)\ny=mx+b$

Hence, for the point (0,-2) and by replacing for dy/dx

$m=(dy)/(dx)_((0,-2))=(4(0)+10(0))/(4(-2)^3-8(-2))=0$

Hence m=0, that is, the tangent line to the point is a horizontal line that cross the y axis for y=-2. The equation is:

y=(0)x+b = -2

HOPE THIS HELPS!!

In order to find the equation of the tangent line to the curve y²(y² - 4) = x²(x² - 5) at the point (0, -2), we will use the method of implicit differentiation. Here are the steps:

Step 1: Differentiate Each Side of the Given Equation with Respect to x

Applying the chain rule to differentiate y²(y² - 4) with respect to x gives:
2y*y'(y² - 4) + y²*2y*y' = d/dx [y²(y² - 4)]
The chain rule is also applied to differentiate x²(x² - 5) with respect to x, yielding:
2x(x² - 5) + x²*2x = d/dx [x²(x² - 5)]

Step 2: Equate the Two Expressions Found from Step 1 and Solve for y'

2y*y'(y² - 4) + y²*2y*y' = 2x(x² - 5) + x²*2x

This equation can be solved by isolating y' (the derivative of y with respect to x), which represents the slope of the tangent line.

Step 3: Use the Given Point (0, -2) to Find the Slope of the Tangent Line

Substitute x = 0 and y = -2 into the equation found in Step 2 to get the specific value for the slope at the given point.

Step 4: Use the Point-Slope Form of the Line to Write the Equation of the Tangent Line

The point-slope form of the line y - y₁ = m(x - x₁) can be used to write the equation of the tangent line. We substitute for x₁ and y₁ with the coordinates of the given point (0, -2), and m with the slope found from Step 3.

The resulting equation represents the tangent line to the curve at the given point (0, -2). Please note that the full calculation may result in a complex slope due to the nature of the given curve equation. Nonetheless, this process illustrates the application of implicit differentiation and the point-slope form of a line in finding the equation of a tangent line to a curve.