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Taya2010 [7]
3 years ago
6

There are 5 red marbles, 4 blue marble, 2 green marbles, and 1 yellow marble in the bag. What is the probability of pulling out

each of the following?
Mathematics
1 answer:
kykrilka [37]3 years ago
3 0

Answer:

Red : 5/12

Blue : 1/3

Green: 1/6

Yellow : 1/12

Step-by-step explanation:

5 red marbles + 4 blue marbles + 2 green marbles + 1 yellow marble = 12 total marbles

5 red marbles / 12 total marbles = 5/12

4 blue marbles / 12 total marbles = 4/12 = 1/3

2 green marbles / 12 total marbles = 2/12 = 1/6

1 yellow  marbles / 12 total marbles = 1/12

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Answer:

3

Step-by-step explanation:

Because in 3xy 3 is the numerical coefficient

3 0
3 years ago
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Determine the amplitude and period of each function without graphing.
Luba_88 [7]

Answer:

Step-by-step explanation:

We know for a function:

y = Asinωt

=> on comparison, <u>A = 5 m</u>

=> T = 2π / ω

=> T = 2π / 2

=> <u>T = π seconds</u>

5 0
2 years ago
What are irregular vowels controlled by w
Mama L [17]

Answer:

It is a vowel only when it teams up with an <a>, <e>, or <o> to spell a single sound—as in the words draw, few, and low. So the letter <w> is a vowel only in the two-letter teams <aw>, <ew>, and <ow>. Everywhere else <w> is a consonant

Step-by-step explanation:

5 0
2 years ago
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Construct a 99% confidthence interval for the population mean .Assume the population has a normal distribution. A group of 19 ra
Kobotan [32]
<h2>Answer with explanation:</h2>

Confidence interval for mean, when population standard deviation is unknown:

\overline{x}\pm t_{\alpha/2}\dfrac{s}{\sqrt{n}}

, where \overline{x} = sample mean

n= sample size

s= sample standard deviation

t_{\alpha/2} = Critical t-value for n-1 degrees of freedom

We assume the population has a normal distribution.

Given, n= 19 , s= 3.8 , \overline{x}=22.4

\alpha=1-0.99=0.01

A) Critical t value for \alpha/2=0.005 and degree of 18 freedom

t_{\alpha/2} = 2.8784

B) Required confidence interval:

22.4\pm ( 2.8744)\dfrac{3.8}{\sqrt{19}}\\\\=22.4\pm2.5058\\\\=(22.4-2.5058,\ 22.4+2.5058)=(19.8942,\ 24.9058)\approx(19.9,\ 24.9)

Lower bound = 19.9 years

Uppen bound = 24.9 years

C) Interpretation: We are 99% confident that the true population mean of lies in (19.9, 24.9) .

5 0
3 years ago
The graph below shows the average Valentine’s Day spending between 2003 and 2012
Lapatulllka [165]
The average rate of change of a graph between two intervals is given by the difference in value of the values on the graph of the two interval divided by the difference between the two intervals.

Part A.

From the graph the average Valentine's day spending in 2005 is 98 while the average Valentine's day spending in 2007 is 120.

The average rate of change in spending between 2005 and 2007 is given by

\frac{120-98}{2007-2005} = \frac{22}{2} =\$11/year



Part B

From the graph the average Valentine's day spending in 2004 is 100 while the average Valentine's day spending in 2010 is 103.

The average rate of change in spending between 2004 and 2010 is given by

\frac{103-100}{2010-2004} = \frac{3}{6} =\$0.5/year



Part C:

From the graph the average Valentine's day spending in 2009 is 102 while the average Valentine's day spending in 2010 is 103.

The average rate of change in spending between 2009 and 2010 is given by

\frac{103-102}{2010-2009} = \$1/year

7 0
3 years ago
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