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Alex777 [14]
1 year ago
5

Suppose you like to keep a jar of change on your desk currently at the jar contains the followingWhat is the probability that yo

u reach into the jar and grab a penny and then without replacement a dime? Express your answer as a fraction or decimal number rounded to four decimal places

Mathematics
1 answer:
Tju [1.3M]1 year ago
7 0
Answer:

The probability of a penny and then without replacement a dime = 7/376

Explanation:

Given:

A jar contains:

6 Pennies, 7 Dimes, 16 Nickels, and 19 Quarters

To find:

the probability when you reach into the jar and grab a penny and then without replacement a dime

Total coins = 6 + 7 + 16 + 19

Total coins = 48

Probability of picking a penny = number of pennies/total coins

Probability of picking a penny = 6/48

Probability of picking a dime after a penny without replacement:

Since we are not replacing the first pick, the total coins will reduce by 1

Total coin for 2nd pick = 48 - 1 = 47

Pr(dime after a penny without replacement) = number of dime/total coin

Pr(dime after a penny without replacement) = 7/47

The probability of a penny and then without replacement a dime = Probability of picking a penny ×

Pr(dime after a penny without replacement)

\begin{gathered} Pr(penny,\text{ then a dime without replacement\rparen= }\frac{6}{48}\times\frac{7}{47} \\  \\ Pr(penny,\text{ then a dime without replacement\rparen= }\frac{1}{8}\times\frac{7}{47} \\  \\ Pr(penny,\text{ then a dime without replacement\rparen= }\frac{7}{376\frac{}{}} \\  \\ Pr(penny,\text{ then a dime without replacement\rparen= 0.0186} \end{gathered}

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x ≤  \frac{-15}{2}

Step-by-step explanation:

So first you would start by getting rid of the -4. To get rid of -4, you would need to + 4 on each side leaving you with -2x ≤ 15. Now you need to get rid of -2 from x so that would convert equation to x ≤  \frac{-15}{2}. Hoped it helped!

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Drag the expressions into the boxes to correctly complete the table.
lora16 [44]

Answer:

SUMMARY:

x^4+\frac{5}{x^3}-\sqrt{x}+8                               →    Not a Polynomial

-x^5+7x-\frac{1}{2}x^2+9                           →    A Polynomial

x^4+x^3\sqrt{7}+2x^2-\frac{\sqrt{3}}{2}x+\pi              →    A Polynomial

\left|x\right|^2+4\sqrt{x}-2                                   →    Not a Polynomial

x^3-4x-3                                        →    A Polynomial

\frac{4}{x^2-4x+3}                                              →    Not a Polynomial

Step-by-step explanation:

The algebraic expressions are said to be the polynomials in one variable which consist of terms in the form ax^n.

Here:

n = non-negative integer

a = is a real number (also the the coefficient of the term).

Lets check whether the Algebraic Expression are polynomials or not.

Given the expression

x^4+\frac{5}{x^3}-\sqrt{x}+8

If an algebraic expression contains a radical in it then it isn’t a polynomial. In the given algebraic expression contains \sqrt{x}, so it is not a polynomial.

Also it contains the term \frac{5}{x^3} which can be written as 5x^{-3}, meaning this algebraic expression really has a negative exponent in it which is not allowed. Therefore, the expression x^4+\frac{5}{x^3}-\sqrt{x}+8 is not a polynomial.

Given the expression

-x^5+7x-\frac{1}{2}x^2+9

This algebraic expression is a polynomial. The degree of a polynomial in one variable is considered to be the largest power in the polynomial. Therefore, the algebraic expression is a polynomial is a polynomial with degree 5.

Given the expression

x^4+x^3\sqrt{7}+2x^2-\frac{\sqrt{3}}{2}x+\pi

in a polynomial with a degree 4. Notice, the coefficient of the term can be in radical. No issue!

Given the expression

\left|x\right|^2+4\sqrt{x}-2

is not a polynomial because algebraic expression contains a radical in it.

Given the expression

x^3-4x-3

a polynomial with a degree 3. As it does not violate any condition as mentioned above.

Given the expression

\frac{4}{x^2-4x+3}

\mathrm{Apply\:exponent\:rule}:\quad \:a^{-b}=\frac{1}{a^b}

Therefore, is not a polynomial because algebraic expression really has a negative exponent in it which is not allowed.

SUMMARY:

x^4+\frac{5}{x^3}-\sqrt{x}+8                               →    Not a Polynomial

-x^5+7x-\frac{1}{2}x^2+9                           →    A Polynomial

x^4+x^3\sqrt{7}+2x^2-\frac{\sqrt{3}}{2}x+\pi              →    A Polynomial

\left|x\right|^2+4\sqrt{x}-2                                   →    Not a Polynomial

x^3-4x-3                                        →    A Polynomial

\frac{4}{x^2-4x+3}                                              →    Not a Polynomial

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