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3 years ago

Ruby uses 1/4 yard of ribbon to make 2 bows.Which expression shows the length of ribbon in each bow?

Mathematics

1 answer:

allochka39001 [22]3 years ago

3 0

What you can do in this case is a rule of three to determine the length of each bow.
We have then:
1/4 ---> 2
x ------> 1
Clearing x we have:
x = (1/2) * (1/4)
x = 1/8
Answer:
the length of ribbon in each bow is
x = 1/8
Equivalently:
x = (1/4) / 2
Option 3

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1 year ago

1. A report from the Secretary of Health and Human Services stated that 70% of single-vehicle traffic fatalities that occur at n

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Using the binomial distribution, it is found that there is a 0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

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Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

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The parameters are:

x is the number of successes.
n is the number of trials.
p is the probability of a success on a single trial.

In this problem:

70% of fatalities involve an intoxicated driver, hence $p = 0.7$ .

A sample of 15 fatalities is taken, hence $n = 15$ .

The probability is:

$P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)$

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$P(X = 11) = C_{15,11}.(0.7)^{11}.(0.3)^{4} = 0.2186$

$P(X = 12) = C_{15,12}.(0.7)^{12}.(0.3)^{3} = 0.1700$

$P(X = 13) = C_{15,13}.(0.7)^{13}.(0.3)^{2} = 0.0916$

$P(X = 14) = C_{15,14}.(0.7)^{14}.(0.3)^{1} = 0.0305$

$P(X = 15) = C_{15,15}.(0.7)^{15}.(0.3)^{0} = 0.0047$

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$P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) = 0.2061 + 0.2186 + 0.1700 + 0.0916 + 0.0305 + 0.0047 = 0.7215$

0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

A similar problem is given at brainly.com/question/24863377

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2 years ago

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