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3 years ago

8

when flipping a coin what is the probability that dave will go to class is 0.7 ,what is the probability that he will not go to c

lass?

1 answer:

Lesechka [4]3 years ago

4 0

0.3 your answer always has to add up to 1 and as there is only two options (heads or tails) the answer must be 1 - 0.7

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7/5=x/3 solve proportion

-BARSIC- [3]

Cross-multiply: 7 * 3 = 5 * x Divide each side by 5 : X = 7 * 3 / 5 = 4.2

6 0

4 years ago

Read 2 more answers

Six customers enter a gift shop during a 3-hour period. On average there are 0.25 customers in the gift shop and each spends 7.5

Stolb23 [73]

Answer:

6 is most directly associated with the flow rate of customers

Step-by-step explanation:

Reason:

0.25 is the number of customers in the gift shop at a given point of time is WIP inventory

7.5 is the average time spent by customer in the shop and it is the throughput time.

Flow rate = WIP inventory/Throughput time

=0.25/7.5

= 0.0333 per minute = 2 per hour or 6 per 3 hour as given in question.

6 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:

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

General Formulas and Concepts:

<u>Calculus</u>

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:

<u>Step 1: Define</u>

<em>Identify given limit</em>.

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

<u>Step 2: Find Limit</u>

Let's start out by <em>directly</em> 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:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \\& = \frac{1}{2\sqrt{3 + 1}} \\& = \boxed{ \frac{1}{4} } \\\end{aligned}$

∴ we have <em>evaluated</em> the given limit.

___

Learn more about limits: brainly.com/question/27807253

Learn more about Calculus: brainly.com/question/27805589

___

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Limits

3 0

2 years ago

Here is the next one jaded :))

Ksju [112]

Answer:

x = 5.56

Step-by-step explanation:

Sin(17) = $\frac{x}{19}$

=> Isolate the "x"

Sin(17) * 19 = x

5.56 = x

Hope this helps!

6 0

4 years ago

A manufacturer wants to double the volume of a 3 in.×2 in.×6 in. 3 i n . × 2 i n . × 6 i n . box, while using as little extra ca

viktelen [127]

Answer: 2 inch dimension will give smallest increase.

Step-by-step explanation:

Length = 3 in

width = 2 in

height = 6 in

Extra cardboard means to find surface area

on doubling the length

length = 6 In

width = 2 In

Height = 6In

Surface area for the above dimensions = 2 [ 6x2+2x6+6x6] = 120 sq in

On doubling the width

length = 3 in

width = 4 in

Height = 6 inch

Surface area for the above dimensions= 2 [ 3x4+4x6+6x3] = 2[54] = 108 sq inches

On doubling height

Length =3 in

width = 2 in

Height = 12 in

Surface area for above dimensions = 2 [ 3x2+2x12+12x3] = 2[6+24+36] = 132 sq inch

On doubling width surface area is minimum.

6 0

3 years ago