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Olegator [25]
3 years ago
10

Which of the following is the correct solution to the linear inequality shown below y 1/3x-2

Mathematics
1 answer:
atroni [7]3 years ago
4 0
We are asked to determine the solution to the linear inequality y <1/3x -2. In this case, we can answer this by equating the equation to zero, transposition and cross multiplication. 

y = 0 = 1/3x-2
2 = 1/3x 
6 = x

If this is the case, x>6
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There are 10 marbles in the bag. There are two purple marbles, 3 red marbles and 5 yellow marbles. You will pick a marble three
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Answer:

30/1000

Step-by-step explanation:

red has a chance of 3/10

purple has a chance of 2/10

yellow has a chance of 5/10

for probability, multiply the denominators together, and the numerators together

3*2*5 = 30

10*10*10 = 1000

So the probability of picking red, then purple, and then yellow with replacement is 30 of 1000

Hope this helps!

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Assume the readings on thermometers are normally distributed with a mean of 0°C and a standard deviation of 1.00°C. Find the pro
lisov135 [29]

Answer:

0.0326 = 3.26% probability that a randomly selected thermometer reads between −2.23 and −1.69.

The sketch is drawn at the end.

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 z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Mean of 0°C and a standard deviation of 1.00°C.

This means that \mu = 0, \sigma = 1

Find the probability that a randomly selected thermometer reads between −2.23 and −1.69

This is the p-value of Z when X = -1.69 subtracted by the p-value of Z when X = -2.23.

X = -1.69

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

Z = \frac{-1.69 - 0}{1}

Z = -1.69

Z = -1.69 has a p-value of 0.0455

X = -2.23

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

Z = \frac{-2.23 - 0}{1}

Z = -2.23

Z = -2.23 has a p-value of 0.0129

0.0455 - 0.0129 = 0.0326

0.0326 = 3.26% probability that a randomly selected thermometer reads between −2.23 and −1.69.

Sketch:

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