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kkurt [141]
3 years ago
8

A motorist traveled 311 miles on 12 gallons of gas. To the nearest tenth, how many miles can the motorist travel on one gallon o

f gas?
Mathematics
2 answers:
inn [45]3 years ago
5 0
The answer is around 25.9 miles
\frac{12}{311}  =  \frac{1}{ x}   \\ x =  \frac{311}{12}  = 25.9




good luck
gizmo_the_mogwai [7]3 years ago
4 0
Hello!

You have to find how far he goes on one gallon

To find this you do

311/12 = 29.1666667

We can round this to 29.2

The answer is 29.2

Hope this helps!
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A survey report states that 70% of adult women visit their doctors for a physical examination at least once in two years. If 20
irakobra [83]

Answer:

a) 0.3921 = 39.21% probability that fewer than 14 of them have had a physical examination in the past two years.

b) 0.107 = 10.7% probability that at least 17 of them have had a physical examination in the past two years.

Step-by-step explanation:

For each women, there are only two possible outcomes. Either they visit their doctors for a physical examination at least once in two years, or they do not. The probability of a woman visiting their doctor at least once in this period is independent of any other women. This means that we use the binomial probability distribution to solve this question.

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.

70% of adult women visit their doctors for a physical examination at least once in two years.

This means that p = 0.7

20 adult women

This means that n = 20

(a) Fewer than 14 of them have had a physical examination in the past two years.

This is:

P(X < 14) = 1 - P(X \geq 14)

In which

P(X \geq 14) = P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)

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

P(X = 14) = C_{20,14}.(0.7)^{14}.(0.3)^{6} = 0.1916

P(X = 15) = C_{20,15}.(0.7)^{15}.(0.3)^{5} = 0.1789

P(X = 16) = C_{20,16}.(0.7)^{16}.(0.3)^{4} = 0.1304

P(X = 17) = C_{20,14}.(0.7)^{17}.(0.3)^{3} = 0.0716

P(X = 18) = C_{20,18}.(0.7)^{18}.(0.3)^{2} = 0.0278

P(X = 19) = C_{20,19}.(0.7)^{19}.(0.3)^{1} = 0.0068

P(X = 20) = C_{20,20}.(0.7)^{20}.(0.3)^{0} = 0.0008

So

P(X \geq 14) = P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) = 0.1916 + 0.1789 + 0.1304 + 0.0716 + 0.0278 + 0.0068 + 0.0008 = 0.6079

P(X < 14) = 1 - P(X \geq 14) = 1 - 0.6079 = 0.3921

0.3921 = 39.21% probability that fewer than 14 of them have had a physical examination in the past two years.

(b) At least 17 of them have had a physical examination in the past two years

P(X \geq 17) = P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)

From the values found in item (a).

P(X \geq 17) = P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) = 0.0716 + 0.0278 + 0.0068 + 0.0008 = 0.107

0.107 = 10.7% probability that at least 17 of them have had a physical examination in the past two years.

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Answer:

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Step-by-step explanation:

7v^2+31v+12

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