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Romashka-Z-Leto [24]
3 years ago
11

4. A bag contains five red marbles, two orange marbles, one yellow marble, and two green marbles. Two marbles are drawn from the

bag.
What is the approximate probability one of the chosen marbles is orange and one of the chosen marbles is green?
A. 0.02222
B. 0.04444
C. 0.08889
D. 0.13333
Mathematics
1 answer:
Vitek1552 [10]3 years ago
4 0

Answer:

  • C. 0.08889

Step-by-step explanation:

<u>Total number of marbles:</u>

  • 5 + 2 + 1 + 2 = 10

<u>There are two possibilities</u>

  • Orange then green or green the orange

<u>The probabilities are:</u>

  • P(o&g) = 2/10*2/9 = 2/45
  • P(g&o) =2/10*2/9 = 2/45

<u>Required probability is:</u>

  • P = 2/45*2 = 4/45 =  0.08889

Correct choice is C

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Answer: 2

Step-by-step explanation: These are similar trapezoids so you take 4 * 5 /10 = 2

7 0
3 years ago
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In science class, Farah uses a graduated cylinder with
FromTheMoon [43]

Answer:

47 mL

Step-by-step explanation:

Note each marble increases the height by 3 mL, and we start with 8 mL, so we multiply 3 by 13, and add it to 8

7 0
3 years ago
Find the area of the figure pictured below.
Tomtit [17]

Answer:

64 square units

Step-by-step explanation:

Split the figure into 2 rectangles of dimensions:

4 × 10

8 × 3

And solve.

(4 × 10) + (8 × 3) = 40 + 24 = 64 square units

See the attached image for a model.

4 0
2 years ago
Need the answer asapp with the working out please thanks :)
patriot [66]
Hi,

The two numbers should be 12 and 30. 12=2x2x3 while 30=2x3x5.

Their HCF is 2x3=6 and their LCM is 2x3x2x5. Because of their HFC, we know that they are both multiple of 6. Also, the question says they both are GREATER than 6, so they can’t be 6 but are 6 times “something”. Thanks to the LCM, we know that “something” is equal to 2 for the first number and to 5 for the second one, the numbers hence being 12 and 30.

I hope this helps. If I was not clear enough or if you’d like further explanation please let me know. Also, English is not my first language, so I’m sorry for any mistakes.
7 0
3 years ago
The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 48,564 miles, with a standard
DerKrebs [107]

Answer:

0.0091 = 0.91% probability that the sample mean would be less than 48,101 miles in a sample of 281 tires if the manager is correct

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 48564, \sigma = 3293, n = 281, s = \frac{3293}{\sqrt{281}} = 196.44

What is the probability that the sample mean would be less than 48,101 miles in a sample of 281 tires if the manager is correct?

This is the pvalue of Z when X = 48101. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{48101 - 48564}{196.44}

Z = -2.36

Z = -2.36 has a pvalue of 0.0091

0.0091 = 0.91% probability that the sample mean would be less than 48,101 miles in a sample of 281 tires if the manager is correct

6 0
3 years ago
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