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3 years ago

Write an equivalent representation of the number 5.46 in word form

Mathematics

1 answer:

DerKrebs [107]3 years ago

3 0

Answer:

five and forty-six hundredths

Step-by-step explanation:

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75.5 is not equivalent because 75 1/2% to a decimal is 0.775

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2 years ago

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What is the product of two odd numbers and a multiple of 2

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3 years ago

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2 years ago

A box contains four $1 and six $5 bills. if three bills are selected at random without replacement, find the probability that al

myrzilka [38]

Answer:

C.1/6

Step-by-step explanation:

Initially the box has four $1 and six $5 bills. The probability of selecting a $5 bill in the first trial would be given as;

(number of $5 bills) / (total number of bills)

= (6)/(4+6) = 3/5

If in the first attempt we actually pick a $5 bill, the number of $5 bills will reduce by one to 5. Now, the probability of picking a $5 bill in the second attempt will be given as;

(new number of $5 bills) / (new total number of bills)

= (5)/(4+5) = 5/9

The new number of $5 bills will now be; 6 - 2 = 4 since we have already picked 2 without replacing them.

Now, the probability of picking a $5 bill in the third attempt will be given as;

(new number of $5 bills) / (new total number of bills)

= (4)/(4+4) = 1/2

Since the three attempts are independent, the probability of picking all three $5 bills is;

3/5 * 5/9 * 1/2 = 1/6

7 0

2 years ago

Find the derivative.

krek1111 [17]

Answer:

$\displaystyle f'(x) = \bigg( \frac{1}{2\sqrt{x}} - \sqrt{x} \bigg)e^\big{-x}$

General Formulas and Concepts:

Algebra I

Terms/Coefficients

Expanding/Factoring

Calculus

Differentiation

Derivatives
Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

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

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle f(x) = \frac{\sqrt{x}}{e^x}$

Step 2: Differentiate

Derivative Rule [Quotient Rule]: $\displaystyle f'(x) = \frac{(\sqrt{x})'e^x - \sqrt{x}(e^x)'}{(e^x)^2}$
Basic Power Rule: $\displaystyle f'(x) = \frac{\frac{e^x}{2\sqrt{x}} - \sqrt{x}(e^x)'}{(e^x)^2}$
Exponential Differentiation: $\displaystyle f'(x) = \frac{\frac{e^x}{2\sqrt{x}} - \sqrt{x}e^x}{(e^x)^2}$
Simplify: $\displaystyle f'(x) = \frac{\frac{e^x}{2\sqrt{x}} - \sqrt{x}e^x}{e^{2x}}$
Rewrite: $\displaystyle f'(x) = \bigg( \frac{e^x}{2\sqrt{x}} - \sqrt{x}e^x \bigg) e^{-2x}$
Factor: $\displaystyle f'(x) = \bigg( \frac{1}{2\sqrt{x}} - \sqrt{x} \bigg)e^\big{-x}$

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

Unit: Differentiation

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2 years ago