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GuDViN [60]
4 years ago
12

Statisticians for a roadside assistance company interviewed 50,000 randomly selected United States (US) households. Of those, 15

,595 reported that they had traveled 50 or more miles from home between December 23 and January 4. If there are 115,000,000 US households, approximately how many of them do the interviews suggest traveled 50 or more miles from home at that time?
Mathematics
1 answer:
I am Lyosha [343]4 years ago
8 0

Answer:

35,868,500 people

Step-by-step explanation:

#We calculate the probability of those who traveled 50+ miles.

-Probability is the likelihood of success. n=50000, p=15595:

P(X\geq 50)=\frac{15595}{50000}\\\\=0.3119

-The US population is 115000000. We determine what fraction of the actual population traveled 50+ miles:

E(X)=np, N=115000000,p=0.3119\\\\=0.3119\times 115000000\\\\\\=35868500

Hence, 35,868,500 people traveled 50+ miles.

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Answer:

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Step-by-step explanation:

We have been given that the work required to stretch a spring 2 ft beyond its natural length is 6 ft-lb. We are asked to find the work needed to stretch the spring 6 in. beyond its natural length.

We can represent our given information as:

6=\int\limits^2_0 {F(x)} \, dx

We will use Hooke's Law to solve our given problem.

F(x)=kx

Substituting this value in our integral, we will get:

6=\int\limits^2_0 {kx} \, dx

Using power rule, we will get:

6=\left[ \frac{kx^2}{2} \right ]^2_0

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We know that 6 inches is equal to 0.5 feet.

Work needed to stretch it beyond 6 inches beyond its natural length would be \int\limits^{0.5}_0 {kx} \, dx =\int\limits^{0.5}_0 {3x} \, dx

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\int\limits^{0.5}_0 {3x} \, dx = \left [\frac{3x^2}{2}\right]^{0.5}_0

\frac{3(0.5)^2}{2}-\frac{3(0)^2}{2}\Rightarrow \frac{3(0.25)}{2}-0=\frac{0.75}{2}=0.375

Therefore, 0.375 feet-lb work is needed to stretch it 6 in. beyond its natural length.

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