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goblinko [34]
3 years ago
12

A student rides his bike to school with an average velocity of 10 miles per hour. It takes him 12 minutes to get to school. Writ

e a function that models the student's distance from home. Use t for time in hours and d for distance. Then identify the domain and range of this situation using interval notation--label each one!
Mathematics
1 answer:
Dovator [93]3 years ago
6 0

Answer:

The function is d=10t

domain is 0 to 0.2 hour.

range is 0 to 2 miles.

Step-by-step explanation:

Given that,

Average velocity = 10 miles/hour

Time = 12 min = 0.2 hour

We need to write a function that models the student's distance from home

Using formula of distance

d=v\times t

Where, v = velocity

t = time

d = distance

Put the value into the formula

d=10t

This is a function.

We know that,

Domain :

Domain shows the time.

So, domain = 0 to 0.2 hour

Range :

Range shows the distance.

The range at t =0,

Put the value of t in the function

d = 10\times0=0

The range at t =0.2 hour

d = 10\times0=0

d=2\ miles

So. range = 0 to 2 miles

Hence, The function is d=10t

domain is 0 to 0.2 hour

range is 0 to 2 miles

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Step-by-step explanation:

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Answer:


Step-by-step explanation:

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16) f(x) = x2 + 3x +5<br>Find the x intercepts ​
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Answer:

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Step-by-step explanation:

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3 years ago
Suppose that X has an exponential distribution with mean equal to 10. Determine the following: a. P(X &gt; 10) b. P(X &gt; 20) c
GrogVix [38]

Answer:

(a) The value of P (X > 10) is 0.3679.

(b) The value of P (X > 20) is 0.1353.

(c) The value of P (X < 30) is 0.9502.

(d) The value of x is 30.

Step-by-step explanation:

The probability density function of an exponential distribution is:

f(x)=\lambda e^{-\lambda x};\ x>0, \lambda>0

The value of E (X) is 10.

The parameter λ is:

\lambda=\frac{1}{E(X)}=\frac{1}{10}=0.10

(a)

Compute the value of P (X > 10) as follows:

P(X>10)=\int\limits^{\infty}_{10} {0.10 e^{-0.10 x}} \, dx \\=0.10\int\limits^{\infty}_{10} { e^{-0.10 x}} \, dx\\=0.10|\frac{e^{-0.10 x}}{-0.10} |^{\infty}_{10}\\=|e^{-0.10 x} |^{\infty}_{10}\\=e^{-0.10\times10}\\=0.3679

Thus, the value of P (X > 10) is 0.3679.

(b)

Compute the value of P (X > 20) as follows:

P(X>20)=\int\limits^{\infty}_{20} {0.10 e^{-0.10 x}} \, dx \\=0.10\int\limits^{\infty}_{20} { e^{-0.10 x}} \, dx\\=0.10|\frac{e^{-0.10 x}}{-0.10} |^{\infty}_{20}\\=|e^{-0.10 x} |^{\infty}_{20}\\=e^{-0.10\times20}\\=0.1353

Thus, the value of P (X > 20) is 0.1353.

(c)

Compute the value of P (X < 30) as follows:

P(X

Thus, the value of P (X < 30) is 0.9502.

(d)

It is given that, P (X < x) = 0.95.

Compute the value of <em>x</em> as follows:

P(X

Take natural log on both sides.

ln(e^{-0.10x})=ln(0.05)\\-0.10x=-2.996\\x=\frac{2.996}{0.10}\\ =29.96\\\approx30

Thus, the value of x is 30.

7 0
3 years ago
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