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3 years ago

The annual health care costs for two new companies that opened in the year 2000 are shown on the graphs below.

Mathematics

2 answers:

Mama L [17]3 years ago

7 0

Company 1:
Average yearly increases in health care costs for the Company 1 over the first 12 years is:
r1=[f(12)-f(0)]/(12-0)
r1=[f(12)-f(0)]/12

f(12)=(24,000+36,000)/2
f(12)=60,000/2
f(12)=30,000

f(0)=12,000

r1=(30,000-12,000)/12
r1=18,000/12
r1=1,500

Company 2:
Average yearly increases in health care costs for the Company 2 over the first 12 years is:
r2=[g(12)-g(0)]/(12-0)
r2=[g(12)-g(0)]/12

g(12)=(36,000+48,000)/2
g(12)=84,000/2
g(12)=42,000

g(0)=[12,000+(12,000+24,000)/2]/2
g(0)=(12,000+36,000/2)/2
g(0)=(12,000+18,000)/2
g(0)=30,000/2
g(0)=15,000

r2=(42,000-15,000)/12
r2=27,000/12
r2=2,250

r2=2,250>1,500=r1
r2-r1=2,250-1500
r2-r1=750

Answer: Fourth option:
Over the first 12 years, the average yearly increase in health care costs for company 2 is greater than the average yearly increase in health care costs for company 1 by about $750.

velikii [3]3 years ago

3 0

Answer:

Answer: Fourth option:

Over the first 12 years, the average yearly increase in health care costs for company 2 is greater than the average yearly increase in health care costs for company 1 by about $750.

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The width at the base of parabolic tunnel is 7 m.

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Solution:

The width at the base of tunnel is 7 m.

Let the graph of the parabola intersect the x-axis at x=0 and x=7. It means x and (x-7) are the factors of the height function.

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Where, a is a constant.

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Putting x=3 and h(x)=4 in (i), we get

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$4=a(3)(-4)$

$\dfrac{4}{(3)(-4)}=a$

$-\dfrac{1}{3}=a$

Putting $a=-\dfrac{1}{3}$ , we get

$h(x)=-\dfrac{1}{3}x(x-7)$ ...(ii)

The center of the parabola is the midpoint of 0 and 7, i.e., 3.

The width of the truck is 4 m. If is passes through the center then the truck must m 2 m on the left side of the center and 2 m on the right side of the center.

2 m on the left side of the center is x=1.5.

A trailer truck carrying cars is 4 m wide and 2.8 m high is going to drive through the tunnel is possible if h(1.5) is greater than 2.8.

Putting x=1.5 in (ii), we get

$h(1.5)=-\dfrac{1}{3}(1.5)(1.5-7)$

$h(1.5)=-(0.5)(-5.5)$

$h(1.5)=2.75$

Since h(1.5)<2.8, therefore the trailer truck carrying cars is 4 m wide and 2.8 m high is going to drive through the tunnel is not possible.

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Answer:

(a) The value of P (None) is 0.062.

(b) The value of P(at least one) is 0.938.

(d) The event is not unusual.

Step-by-step explanation:

Let X = number of households watching the show.

The probability of the random variable x is, P (X) = p = 0.18.

The sample selected for the survey is of size, n = 14

The random variable X follows a Binomial distribution with parameter n = 14 and p = 0.18.

The probability of a Binomial distribution is computed using the formula:

$P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0,1,2,...$

(a)

Compute the probability that none of the households are tuned to CSI: Shoboygan as follows:

$P(X=0)={14\choose 0}(0.18)^{0}(1-0.18)^{14-0}=1\times1\times0.06214=0.062$

Thus, the value of P (None) is 0.062.

(b)

Compute the probability that at least one household is tuned to CSI: Shoboygan as follows:

P (X ≥ 1) = 1 - P (X < 1)

= 1 - P (X = 0)

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Thus, the value of P(at least one) is 0.938.

(c)

Compute the probability that at most one household is tuned to CSI: Shoboygan as follows:

P (X ≤ 1) = P (X = 0) + P (X = 1)

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Thus, the value of P(at most one) is 0.253.

(d)

An event that has a very low probability of occurrence is known as an unusual event.

The probability of the event "at most one household is tuned to CSI: Shoboygan" is 0.253.

This probability value is not low.

Hence, the event is not unusual.

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