All categories

1 year ago

Explain the tangent line problem

1 answer:

8 0

The Tangent Line Problem 1/3How do you find the slope of the tangent line to a function at a point Q when you only have that one point? This Demonstration shows that a secant line can be used to approximate the tangent line. The secant line PQ connects the point of tangency to another point P on the graph of the function. As the distance between the two points decreases, the secant line becomes closer to the tangent line.

You might be interested in

Write an expression that represents the volume of a rectangular prism (box) with length 2003-02-02-00-00_files/i0260000.jpg, wid

k0ka [10]

h=5 in

w-6 in

l=12 in

SA/V=2*12*6+2*6*5+2*12*5/12*6*5

817:180

This is just an example do not use this exact equation and number! Hope it helps. : )

3 0

1 year ago

A supervisor finds the mean number of miles that the employees in a department live from work. Which mileage is within a z-score

Vikentia [17]

The z-score tells you how many standard deviations from the mean.

1.5 * 3.6 = 5.4 miles 

So anything within 5.4 miles of the average (29). 

The range would be: 
29 - 5.4 = 23.6 
to: 
29 + 5.4 = 34.4 

23.6 ≤ x ≤ 34.4 

Answer: 
B) 24 miles

7 0

2 years ago

Read 2 more answers

A construction company is considering submitting bids for two contracts. It will cost the company \$10{,}000$10,000dollar sign,

baherus [9]

Answer:

0 dollars

=E(M)

=μ

=−$10,000(0.81)+$40,000(0.18)+$90,000(0.01)

=−8,100+7,200+900

8 0

2 years ago

Read 2 more answers

The orbit of the planet Venus is nearly circular. An astronomer develops a model for the orbit in which the sun has coordinates

klio [65]

Since the Venus orbits round the sun, the sun is the center of the circular path of the revolution of the planet, Venus.

Thus, the distance of the planet, Venus fron the sun is given by the distance between the points (0, 0) and (41, 53).

Recall that the distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $d= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Thus, the distance between the points (0, 0) and (41, 53) is given by:

$d= \sqrt{(41-0)^2+(53-0)^2} \\ \\ = \sqrt{41^2+53^2} = \sqrt{1,681+2,809} \\ \\ = \sqrt{4,490} =67 \ units$

Given that each unit of the plane represents 1 million miles, therefore, the distance from the sun to the Venus is 67 million miles.

8 0

1 year ago

The average credit card debt for a recent year was $8,776. Five years earlier the average credit card debt was $8,189. Assume sa

Anestetic [448]

Answer:

We reject H₀. We support that the new average credit card debt is bigger than the previous average

Step-by-step explanation:

Five years earlier

μ = 8189

σ = 690

Sample size n = 32

Recent year debt

x = 8776

Sample size n = 32

a) Hypothesis Test:

Null Hypothesis H₀ x = μ = 8189

Alternative Hypothesis Hₐ x > μ

b) z(c) Alternative Hypothesis establishes that the test is a one tail-test to the right.

z(c) for significance level α = 0.05 is from z-table z(c) = 1,64

c) z(s) = ( x - μ ) / σ /√n

z(s) = ( 8776 - 8189 ) / 690 /√32

z(s) = 587 *5,66/ 690

z(s) = 4,81

d) Comparing z(c) and z(s)

z(s) > z(c) Then z(c) is in the rejection region and we reject H₀

e) We have evidence that at 95 % of confidence the new value for the debt in credit card is now bigger than the average

7 0

1 year ago