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2 years ago

PLEASE HELP ME I NEED ANSWERS FAST!!! No links plz

Mathematics

1 answer:

poizon [28]2 years ago

4 0

Answer:

$220,000 I am pretty sure this is correct!

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The lengths of the diagonals of rectangle ABCD intersect at E. If AE =x+4 and CE= 3x-12. What is the length of BD?

swat32

Answer:

DB = 24

Step-by-step explanation:

First, note that the diagonals of a rectangle are equal and bisect each other. In other words, DB = CA and CE = EA and DE = BE.

Also, AE + CE = CA

So, using this, we can write this equation:

AE = CE

x + 4 = 3x -12

Subtract 4 from both sides.

x = 3x -16

Subtract 3x from both sides.

-2x = -16

Divide both sides by -2

x = 8

Then, substitute this into AE + CE = CA

x + 4 + 3x - 12 =

8 + 4 + 24 - 12 = 24

Then, because CA = DB,

DB = 24

I hope this helps! Feel free to ask any questions! :)

5 0

2 years ago

Read 2 more answers

It is assumed that the mean weight of a Labrador retriever is 70 pounds. A breeder claims that the average weight of an adult ma

Sindrei [870]

Answer:

z(s) is in the acceptance region we accept H₀. We don´t have enough evidence to support the breeder´s claim

Step-by-step explanation:

We will test the breeder´s claim at 95% ( CI) or significance level

α = 5 % α = 0,05 α /2 = 0,025

Sample Information:

sample size n = 45

sample mean x = 72,5 pounds

Sample standard deviation s = 16,1

1.-Hypothesis Test:

Null Hypothesis H₀ x = 70

Alternative Hypothesis Hₐ x ≠ 70

Alternative hypothesis contains the information about what kind of test has to be developed ( in this case it will be a two-tail tets)

2.-z (c) is from z-table z(c) = 1,96

3.- z(s) = ( x - 70 ) / 16,1 / √45

z(s) = (72,5 -70 ) *√45 / 16,1

z(s) = 2,5 * 6,71 / 16,1

z(s) = 1,04

4.-Comparing z(s) and z(c)

z(s) < z(c)

Then z(s) is in the acceptance region we accept H₀. We don´t have enough evidence to support the breeder´s claim

7 0

3 years ago

Read 2 more answers

Broccoli costs $1.50 PER POUND at a grocery store. How much money does 64 OUNCES of broccoli cost?

Gnoma [55]

$6.00
Show your work in the attachment. Down below

5 0

3 years ago

What is Limit of StartFraction StartRoot x + 1 EndRoot minus 2 Over x minus 3 EndFraction as x approaches 3?

scoray [572]

Answer:

$\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} = \boxed{ \frac{1}{4} }$

General Formulas and Concepts:

Calculus

Limits

Limit Rule [Variable Direct Substitution]:
$\displaystyle \lim_{x \to c} x = c$

Special Limit Rule [L’Hopital’s Rule]:
$\displaystyle \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$

Differentiation

Derivatives
Derivative Notation

Derivative Property [Addition/Subtraction]:
$\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$
Derivative Rule [Basic Power Rule]:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]:
$\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

Step 1: Define

Identify given limit.

$\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3}$

Step 2: Find Limit

Let's start out by directly evaluating the limit:

[Limit] Apply Limit Rule [Variable Direct Substitution]:
$\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} = \frac{\sqrt{3 + 1} - 2}{3 - 3}$
Evaluate:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \frac{\sqrt{3 + 1} - 2}{3 - 3} \\& = \frac{0}{0} \leftarrow \\\end{aligned}$

When we do evaluate the limit directly, we end up with an indeterminant form. We can now use L' Hopital's Rule to simply the limit:

[Limit] Apply Limit Rule [L' Hopital's Rule]:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\\end{aligned}$
[Limit] Differentiate [Derivative Rules and Properties]:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \leftarrow \\\end{aligned}$
[Limit] Apply Limit Rule [Variable Direct Substitution]:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \\& = \frac{1}{2\sqrt{3 + 1}} \leftarrow \\\end{aligned}$
Evaluate:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \\& = \frac{1}{2\sqrt{3 + 1}} \\& = \boxed{ \frac{1}{4} } \\\end{aligned}$