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3 years ago

9

8. A banker's loan officer rates applications for credit. The rates are normally distributed with a mean of 200 and a standard d

eviation of 65. If 12 applications are randomly selected

1 answer:

masha68 [24]3 years ago

3 0

Answer:

802

Step-by-step explanation:

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I NEED THIS ASAP please. can someone please answer this question i put the picture its on Simplify Expressions im giving 15 poin

kakasveta [241]

Answer:

in step 4

Step-by-step explanation:

we have

$0.5(-12c+6)-3(c+4)+10(c-5)\\\\step1=-12c(0.5)+6(0.5)-3(c+4)+10(c-5)\\\\=-6c+3-3(c+4)+10(c-5)\\\\$

so step 1 is correct

step 2

$-6c+3-3(c+4)+10(c-5)\\\\=-6c+3+c(-3)+4(-3)+10(c-5)\\\\=-6c+3-3c-12+10(c-5)$

so step 2 is correct

step 3

$-6c+3-3c-12+10(c-5)\\\\=-6c+3-3c-12+c(10)-5(10)\\\\=-6c+3-3c-12+10c-50$

so step 3 is correct

step 4

$-6c+3-3c-12+10c-50\\\\=(-6c-3c+10c)+(3-12-50)\\\\=(-6-3+10)c+(-9-50)\\\\=(-9+10)c-59\\\\=c-59$

so step 4 is incorrect

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4 years ago

Kitty [74]

Answer:

2

Step-by-step explanation:

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3 years ago

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masha68 [24]

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3 years ago

PLEASE HELP !!!

olga55 [171]

Answer:

B: x=-2

Step-by-step explanation:

This is because x=-2 is where the parabola is split into two equal halves.

5 0

3 years ago

Implicit differentiation Please help

Anvisha [2.4K]

Answer:

$y''(-1) =8$

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

<u>Algebra I</u>

Factoring

<u>Calculus</u>

Implicit Differentiation

The derivative of a constant is equal to 0

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Product Rule: $\frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)$

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

Quotient Rule: $\frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}$

Step-by-step explanation:

<u>Step 1: Define</u>

-xy - 2y = -4

Rate of change of the tangent line at point (-1, 4)

<u>Step 2: Differentiate Pt. 1</u>

<em>Find 1st Derivative</em>

Implicit Differentiation [Product Rule/Basic Power Rule]: $-y - xy' - 2y' = 0$
[Algebra] Isolate <em>y'</em> terms: $-xy' - 2y' = y$
[Algebra] Factor <em>y'</em>: $y'(-x - 2) = y$
[Algebra] Isolate <em>y'</em>: $y' = \frac{y}{-x-2}$
[Algebra] Rewrite: $y' = \frac{-y}{x+2}$

<u>Step 3: Find </u><em><u>y</u></em>

Define equation: $-xy - 2y = -4$
Factor <em>y</em>: $y(-x - 2) = -4$
Isolate <em>y</em>: $y = \frac{-4}{-x-2}$
Simplify: $y = \frac{4}{x+2}$

<u>Step 4: Rewrite 1st Derivative</u>

[Algebra] Substitute in <em>y</em>: $y' = \frac{-\frac{4}{x+2} }{x+2}$
[Algebra] Simplify: $y' = \frac{-4}{(x+2)^2}$

<u>Step 5: Differentiate Pt. 2</u>

<em>Find 2nd Derivative</em>

Differentiate [Quotient Rule/Basic Power Rule]: $y'' = \frac{0(x+2)^2 - 8 \cdot 2(x + 2) \cdot 1}{[(x + 2)^2]^2}$
[Derivative] Simplify: $y'' = \frac{8}{(x+2)^3}$

<u>Step 6: Find Slope at Given Point</u>

[Algebra] Substitute in <em>x</em>: $y''(-1) = \frac{8}{(-1+2)^3}$
[Algebra] Evaluate: $y''(-1) =8$

6 0

3 years ago

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