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exis [7]
1 year ago
12

marcus has a spinner with 3 red sections, 2blue sections, and 1 purple section match the event of landing on each color to the c

orrect probability
Mathematics
1 answer:
wolverine [178]1 year ago
8 0

Answer:

see below

Step-by-step explanation:

3 red sections, 2blue sections, and 1 purple section = 6 sections

P( red) = red/total = 3/6 =1/2

P( blue) = blue/total = 2/6 =1/3

P( purple) = purple/total = 1/6

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Answer:

14.5 =x

Step-by-step explanation:

Since this is a right triangle we can use trig functions

sin theta = opposite side/ hypotenuse

sin A = BC / AB

sin 65 = x /16

Multiply each side by 16

16 sin 65 =x

14.50092459 =x

14.5 =x

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Step-by-step explanation:

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A cylinder has its height doubled and its radius cut to 1/3. What is the ratio of the volumes of the modified cylinder to the or
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Helpmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm
alexandr402 [8]

Answer:

  B.  1.2×10^-3 mm

Step-by-step explanation:

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  1.2×10^-3 mm

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6 0
2 years ago
According to a human modeling​ project, the distribution of foot lengths of women is approximately Normal with a mean of 23.3 ce
Yakvenalex [24]

Answer:

26.11% of women in the United States will wear a size 6 or​ smaller

Step-by-step explanation:

Problems of normally distributed samples are 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.

In this problem, we have that:

\mu = 23.3, \sigma = 1.4

In the United​ States, a​ woman's shoe size of 6 fits feet that are 22.4 centimeters long. What percentage of women in the United States will wear a size 6 or​ smaller?

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