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4 years ago

Could someone please solve these for me? Thanks in advance!

Mathematics

1 answer:

Amiraneli [1.4K]4 years ago

8 0

She spent $9 on the taxi

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3 years ago

Read 2 more answers

A box contains 5 red balls, 6 white balls and 9 black balls. Two balls are drawn at

valina [46]

Answer:

$P(Same)=\frac{61}{190}$

Step-by-step explanation:

Given

$Red = 5$

$White = 6$

$Black = 9$

Required

The probability of selecting 2 same colors when the first is not replaced

The total number of ball is:

$Total = 5 + 6 + 9$

$Total = 20$

This is calculated as:

$P(Same)=P(Red\ and\ Red) + P(White\ and\ White) + P(Black\ and\ Black)$

So, we have:

$P(Same)=\frac{n(Red)}{Total} * \frac{n(Red) - 1}{Total - 1} + \frac{n(White)}{Total} * \frac{n(White) - 1}{Total - 1} + \frac{n(Black)}{Total} * \frac{n(Black) - 1}{Total - 1}$

Note that: 1 is subtracted because it is a probability without replacement

$P(Same)=\frac{5}{20} * \frac{5 - 1}{20- 1} + \frac{6}{20} * \frac{6 - 1}{20- 1} + \frac{9}{20} * \frac{9- 1}{20- 1}$

$P(Same)=\frac{5}{20} * \frac{4}{19} + \frac{6}{20} * \frac{5}{19} + \frac{9}{20} * \frac{8}{19}$

$P(Same)=\frac{20}{380} + \frac{30}{380} + \frac{72}{380}$

$P(Same)=\frac{20+30+72}{380}$

$P(Same)=\frac{122}{380}$

$P(Same)=\frac{61}{190}$

4 0

3 years ago

If if the branding division were to experience the same percentage increase from year 2 to year 3 as they did from year 1 to yea

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Since the basis is from year 1 to year 2, calculate first for the difference of their percentages. That would be:

Difference = year 2 - year 1
Difference = 2.32% - 1.1% = 1.22%

We apply this same value of percentage increase from year 2 to year. Thus, the percentage for year 3 is:

% Year 3 = % Year 2 + percentage increase
% Year 3 = 2.32% + 1.22%
% Year 3 = 3.54%

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3 years ago

In a certain test, the number of successful candidates was three times than that of unsuccessful candidate, if there had been 16

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Answer:

The number of candidates is 136.

Step-by-step explanation:

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3 years ago