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STatiana [176]
2 years ago
12

One ball is selected at random from a bag containing 12 balls of which x are white

Mathematics
1 answer:
damaskus [11]2 years ago
7 0

Answer:

a) 12/18

b) 6

Step-by-step explanation:

If adding 6 white balls doubles the probability then the original amount of balls is 6 and with the new amount of 18, the likelihood of picking a white ball becomes 12/18.

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Mark has downloaded four times as many songs on his music player as Chloe. If mark has 440 songs,how many songs does Chloe have?
VladimirAG [237]
440/4 = 110

Chloe has 110 songs.
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3 years ago
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7+(a+b)=(7+a)+b please help asap
elena-s [515]

Answer:

We are using the Associate property of addition

Step-by-step explanation:

7+(a+b)=(7+a)+b

We are using the Associate property of addition

5 0
4 years ago
Which inequality is true?
11Alexandr11 [23.1K]

Answer:

9\pi >27

Step-by-step explanation:

9\pi >27

Divide both sides by 9.

9\pi/9 >27/9

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3.14159265358.. >3

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3 years ago
HELP WILL MARK BRAINLIEST
kakasveta [241]

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Step-by-step explanation:

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6 0
3 years ago
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A chocolate bar is produced with an advertised mass of 200 g.
Elan Coil [88]

Answer:

The proportion of the chocolate bars produced that have mass within one standard deviation of the mean is 0.6826 = 68.26%.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions 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.

A consumer organization finds that the mass of the chocolate bar is normally distributed with a mean of 200.4 g and a standard deviation of 0.05 g.

This means that \mu = 200.4, \sigma = 0.05

a) Find the proportion of the chocolate bars produced that have mass within one standard deviation of the mean.

pvalue of Z when X = 200.4 + 0.05 = 200.9 subtracted by the pvalue of Z when X = 200.4 - 0.05 = 199.9.

X = 200.9

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

Z = \frac{200.9 - 200.4}{0.05}

Z = 1

Z = 1 has a pvalue of 0.8413

X = 199.9

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

Z = \frac{199.9 - 200.4}{0.05}

Z = -1

Z = -1 has a pvalue of 0.1587

0.8413 - 0.1587 = 0.6826

The proportion of the chocolate bars produced that have mass within one standard deviation of the mean is 0.6826 = 68.26%.

8 0
3 years ago
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