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Gnom [1K]
4 years ago
7

PLEASE HELP ME WITH THIS & IT DUE tomorrow.

Mathematics
1 answer:
sattari [20]4 years ago
4 0

Answer:

the first one is 2.5, because if you divide the blue in half, you need to divide the yellow in half too

Answer 2:

copy and paste what i write or something:

3+2+1+1= 1 plum purple gallon. So for 3 gallons, you multiply each by 3.

So, Red tint=9

     Blue tint=6

     Yellow tint=3 and

     White tint=3

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Given that P = (-5, 11) and Q = (-6, 4), find the component form and magnitude of vector QP.
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A researcher tests the braking distances of several cars. The braking distance from 60 miles per hour to a complete stop on dry
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Answer:

The longest braking distance one of these cars could have and still be in the bottom 1% is of 116.94 feet.

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.

The braking distances of a sample of cars are normally distributed, with a mean of 129 feet and a standard deviation of 5.18 feet.

This means that \mu = 129, \sigma = 5.18

What is the longest braking distance one of these cars could have and still be in the bottom 1%?

This is the 1st percentile, which is X when Z has a pvalue of 0.01, so X when Z = -2.327.

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

-2.327 = \frac{X - 129}{5.18}

X - 129 = -2.327*5.18

X = 116.94

The longest braking distance one of these cars could have and still be in the bottom 1% is of 116.94 feet.

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