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dybincka [34]
3 years ago
5

Tiana has already taken 1 page of notes on her own, and she will take 1 page during each hour of class. In all, how many hours w

ill Tiana have to spend in class before she will have a total of 43 pages of notes in her notebook?
Mathematics
1 answer:
Novosadov [1.4K]3 years ago
4 0

Answer:

<u>42 Hours.</u>

Step-by-step explanation:

As an equation where y is the total pages and x is hours of class:

y=1+1x since she started with 1 page and does 1 page per hour.

Setting the total pages she needs as 43 (y=43) we get:

43=1+1x

Subtract one on both sides

42=1x

The 1 doesn't need to be written so

x=42

Therefore, it'll take her 42 hours to write all those notes.

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THE SUM 10 INFINITY OF A GEOMETRIC SERIES IS 100. FIND THE TERM IF THE COMMON FIRST RATIO IS - 1/2​
zhuklara [117]

Answer:

a = 150

Step-by-step explanation:

<u>Sum to infinity of a geometric series</u>

S_{\infty}=\dfrac{a}{1-r}\:\textsf{ for }|r| < 1

where:

  • a is the first term
  • r is the common ratio

Given:

  • S_{\infty}=100
  • r=-\dfrac{1}{2}

Substitute the given values into the formula and solve for a:

\begin{aligned}S_{\infty} & =\dfrac{a}{1-r}\\\\\implies 100 & =\dfrac{a}{1-\left(-\frac{1}{2}\right)}\\\\ 100 & =\dfrac{a}{1+\frac{1}{2}}\\\\100 & = \dfrac{a}{\frac{3}{2}}\\\\a & = \dfrac{3}{2} \cdot 100\\\\a & = 150\end{aligned}

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1 year ago
Francisco purchased a 2010 model sedan for $24,999. The dealership offered him a $199/month
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What is 5(6\36+12-3)?<br> (No answer choice)
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Suppose that two cards are randomly selected from a standard​ 52-card deck. ​(a) What is the probability that the first card is
ser-zykov [4K]

Answer:

(a)\frac{1}{17} (b) \frac{1}{16}

Step-by-step explanation:

GIVEN: Suppose that two cards are randomly selected from a standard​ 52 card deck.

TO FIND: (a) What is the probability that the first card is a club and the second card is a club if the sampling is done without​ replacement? ​(b) What is the probability that the first card is a club and the second card is a club if the sampling is done with​ replacement.

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(a)

Probability that first card is club P(A)=\frac{\text{total club cards}}{\text{total cards}}

                                                   =\frac{13}{52}

                                                   =\frac{1}{4}

As sampling is done without replacement.

probability that second card is club  P(B)=\frac{\text{total club cards}}{\text{total cards}}

                                                            =\frac{12}{51}

                                                            =\frac{4}{17}

Probability that first card is club and second card is club =P(A)\times P(B)

                                                                                             =\frac{1}{4}\times\frac{4}{17}=\frac{1}{17}

(b)

Probability that first card is club P(A)=\frac{\text{total club cards}}{\text{total cards}}

                                                   =\frac{13}{52}

                                                   =\frac{1}{4}

As sampling is done with replacement.

probability that second card is club  P(B)=\frac{\text{total club cards}}{\text{total cards}}

                                                            =\frac{13}{52}

                                                            =\frac{1}{4}

Probability that first card is club and second card is club =P(A)\times P(B)

                                                                                             =\frac{1}{4}\times\frac{1}{4}=\frac{1}{16}

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Answer:

175262

Step-by-step explanation:

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