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3 years ago

Find each probability below -- as a fraction, decimal and percent Will give brainliest

Mathematics

2 answers:

Komok [63]3 years ago

7 0

Answer:

I was not sure to what decimal point to round it to

a) Fraction: 2/7 Decimal: 0.28571 Percent: 28.571 %

b) Fraction: 5/7 Decimal: 0.71428 Percent: 71.428 %

c) Fraction: 1/7 Decimal: 0.14285 Percent: 14.285 %

Naily [24]3 years ago

3 0

Answer:

a.)

Fraction: 2/7

Decimal: 0.29 (rounded to nearest tenth)

Percent: 28.57%

b.)

Fraction: 5/7

Decimal: 0.7143

Percent: 71.43%

c.)

Fraction: 1/7

Decimal: 0.143 (rounded to nearest hundredth)

Percent: 14.29% (rounded to nearest tenth)

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What would you do when ok so he said yes would go? 10x2=? 10x11=?

Alla [95]

Answer:

10 x 2 = 20

10 x 11 = 110

Step-by-step explanation:

U can do 10 + 10 = 20

and

U can do 10+10+10+10+10+10+10+10+10+10+10 = 110.

Hope this helps. And what is the other question u are asking about?

5 0

3 years ago

All please Distributive property not multiplication

blagie [28]

You just multiply them normally because there are only two numbers

6 0

3 years ago

Use the equation to answer the following question y=(x-3(x+2)/(x+4)(x-4)(x+2)

givi [52]

Answer:

See below

Step-by-step explanation:

The given rational function is;

$y=\frac{(x-3)(x+2)}{(x+4)(x-4)(x+2)}$

The given function is not continuous where the denominator is equal to zero.

$(x+4)(x-4)(x+2)=0$

The function is discontinuous at $x=-4,x=4,x=-2$

b) The point at x=-2 is a removable discontinuity(hole) because (x+2) is common to both the numerator and the denominator.

The point at x=-4 and x=4 are non-removable discontinuities(vertical asymptotes)

c) The equation of the vertical asymptotes are x=-4 and x=4

To find the equation of the horizontal asymptote, we take limit to infinity.

$\lim_{x\to \infty}\frac{(x-3)(x+2)}{(x+4)(x-4)(x+2)}=0$

The horizontal asymptote is y=0

8 0

3 years ago

6.28 x^2 + 43.96 x + 62.8 + 12.56 x + 25.12

Anuta_ua [19.1K]

The correct answer is 157

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4 years ago

Find the area of the region enclosed by the graphs of these equations. (CALCULUS HELP)

sergiy2304 [10]

Answer:

$\displaystyle A = \frac{20\sqrt{15}}{3}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality

Algebra I

Terms/Coefficients
Graphing
Exponential Rule [Root Rewrite]: $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$

Calculus

Derivatives

Derivative Notation

Derivative of a constant is 0

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Area - Integrals

U-Substitution

Integration Rule [Reverse Power Rule]: $\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C$

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

Integration Property [Addition/Subtraction]: $\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx$

Integration Rule [Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)$

Area of a Region Formula: $\displaystyle A = \int\limits^b_a {[f(x) - g(x)]} \, dx$

Step-by-step explanation:

Step 1: Define

F: y = √(15 - x)

G: y = √(15 - 3x)

H: y = 0

Step 2: Find Bounds of Integration

Solve each equation for the x-value for our bounds of integration.

Set y = 0: 0 = √(15 - x)
[Equality Property] Square both sides: 0 = 15 - x
[Subtraction Property of Equality] Isolate x term: -x = -15
[Division Property of Equality] Isolate x: x = 15

Set y = 0: 0 = √(15 - 3x)
[Equality Property] Square both sides: 0 = 15 - 3x
[Subtraction Property of Equality] Isolate x term: -3x = -15
[Division Property of Equality] Isolate x: x = 5

This tells us that our bounds of integration for function F is from 0 to 15 and our bounds of integration for function G is 0 to 5.

We see that we need to subtract function G from function F to get our area of the region (See attachment graph for visual).

Step 3: Find Area of Region

Integration Part 1

Rewrite Area of Region Formula [Integration Property - Subtraction]: $\displaystyle A = \int\limits^b_a {f(x)} \, dx - \int\limits^d_c {g(x)} \, dx$
[Integral] Substitute in variables and limits [Area of Region Formula]: $\displaystyle A = \int\limits^{15}_0 {\sqrt{15 - x}} \, dx - \int\limits^5_0 {\sqrt{15 - 3x}} \, dx$
[Area] [Integral] Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle A = \int\limits^{15}_0 {(15 - x)^{\frac{1}{2}}} \, dx - \int\limits^5_0 {(15 - 3x)^{\frac{1}{2}}} \, dx$

Step 4: Identify Variables

Set variables for u-substitution for both integrals.

Integral 1:

u = 15 - x

du = -dx

Integral 2:

z = 15 - 3x

dz = -3dx

Step 5: Find Area of Region

Integration Part 2

[Area] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle A = -\int\limits^{15}_0 {-(15 - x)^{\frac{1}{2}}} \, dx + \frac{1}{3}\int\limits^5_0 {-3(15 - 3x)^{\frac{1}{2}}} \, dx$
[Area] U-Substitution: $\displaystyle A = -\int\limits^0_{15} {u^{\frac{1}{2}}} \, du + \frac{1}{3}\int\limits^0_{15} {z^{\frac{1}{2}}} \, dz$
[Area] Reverse Power Rule: $\displaystyle A = -(\frac{2u^{\frac{3}{2}}}{3}) \bigg|\limit^0_{15} + \frac{1}{3}(\frac{2z^{\frac{3}{2}}}{3}) \bigg|\limit^0_{15}$
[Area] Evaluate [Integration Rule - FTC 1]: $\displaystyle A = -(-10\sqrt{15}) + \frac{1}{3}(-10\sqrt{15})$
[Area] Multiply: $\displaystyle A = 10\sqrt{15} + \frac{-10\sqrt{15}}{3}$
[Area] Add: $\displaystyle A = \frac{20\sqrt{15}}{3}$

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

Unit: Area Under the Curve - Area of a Region (Integration)

Book: College Calculus 10e

3 0

3 years ago