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Neporo4naja [7]
3 years ago
5

Which estimation could be used to make sure the answer is reasonable? Sofia mixed 3.15 L of grape juice with 4.75 L of cranberry

juice. Her friends drank 3.95 L of the juice mixture. How much juice was left over?
A.
Answer: The amount left over was 4.05 L.
Estimation: 3.15 + 4.75 is about 7.
7 L – 4 L = 3 L
The amount left over was about 3 L.

B.
Answer: The amount left over was 3.95 L.
Estimation: 3.15 + 4.75 is about 8.
8 L – 4 L = 4 L
The amount left over was a little less than 4 L.

C.
Answer: The amount left over was 11.85 L.
Estimation: 3.15 + 4.75 is about 8 L.
8 L + 4 L = 12 L.
The amount left over was a little less than 12 L.
Mathematics
2 answers:
dangina [55]3 years ago
8 0
Let's go through each answer. 

<span>A.
Answer: The amount left over was 4.05 L.
Estimation: 3.15 + 4.75 is about 7.
7 L – 4 L = 3 L
The amount left over was about 3 L.

This is incorrect because 3.15+4.75 is closer to 8 than 7.

</span><span>B.
Answer: The amount left over was 3.95 L.
Estimation: 3.15 + 4.75 is about 8.
8 L – 4 L = 4 L
The amount left over was a little less than 4 L.

This is correct because 3.15+4.75 is close to 8 and to find the amount left over you do have to subtract. 

The correct answer is </span><span>B.
Answer: The amount left over was 3.95 L.
Estimation: 3.15 + 4.75 is about 8.
8 L – 4 L = 4 L
<span>The amount left over was a little less than 4 L.

Hope this helps! Let me know if you need any more help by commenting or DMing me. If it helped, please rate, thank, and if it really helped, give brainliest. It helps me rank up. Thanks! :)
</span></span>
nataly862011 [7]3 years ago
3 0
I did this too and i got B because 3.15 + 4.75 is 7.90 or you could say about 8
because  8 is .10 more but it is b
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A highway traffic condition during a blizzard is hazardous. Suppose one traffic accident is expected to occur in each 60 miles o
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a) 0.3408 = 34.08% probability that at least one accident will occur on a blizzard day on a 25 mile long stretch of highway.

b) 0.0879 = 8.79% probability that two out of these six blizzard days have no accident on a 25 mile long stretch of highway.

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Step-by-step explanation:

We have the mean for a distance, which means that the Poisson distribution is used to solve this question. For item b, the binomial distribution is used, as for each blizzard day, the probability of an accident will be the same.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given interval.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

Suppose one traffic accident is expected to occur in each 60 miles of highway blizzard day.

This means that \mu = \frac{n}{60}, in which 60 is the number of miles.

(a) What is the probability that at least one accident will occur on a blizzard day on a 25 mile long stretch of highway?

n = 25, and thus, \mu = \frac{25}{60} = 0.4167

This probability is:

P(X \geq 1) = 1 - P(X = 0)

In which

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-0.4167}*(0.4167)^{0}}{(0)!} = 0.6592

P(X \geq 1) = 1 - P(X = 0) = 1 - 0.6592 = 0.3408

0.3408 = 34.08% probability that at least one accident will occur on a blizzard day on a 25 mile long stretch of highway.

(b) Suppose there are six blizzard days this winter. What is the probability that two out of these six blizzard days have no accident on a 25 mile long stretch of highway?

Binomial distribution.

6 blizzard days means that n = 6

Each of these days, 0.6592 probability of no accident on this stretch, which means that p = 0.6592.

This probability is P(X = 2). So

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

P(X = 2) = C_{6,2}.(0.6592)^{2}.(0.3408)^{4} = 0.0879

0.0879 = 8.79% probability that two out of these six blizzard days have no accident on a 25 mile long stretch of highway.

(c) If the probability of damage requiring an insurance claim per accident is 60%, what is the probability of no damaging accidents requiring a insurance claims on an 80 mile stretch of highway?

Probability of damage requiring insurance claim per accident is of 60%, which means that the mean is now:

\mu = 0.6\frac{n}{60} = 0.01n

80 miles:

This means that n = 80. So

\mu = 0.01(80) = 0.8

The probability of no damaging accidents is P(X = 0). So

P(X = 0) = \frac{e^{-0.8}*(0.8)^{0}}{(0)!} = 0.4493

0.4493 = 44.93% probability of no damaging accidents requiring a insurance claims on an 80 mile stretch of highway.

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3 years ago
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