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2 years ago

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Tim's phone service charges $24.25 plus an additional $0.18 for each text message sent per month. If Tim's phone bill was $28.75

, which equation could be used to find how many text messages, x, Tim sent last month?

1 answer:

valina [46]2 years ago

8 0

24.25+ .18 = 28.78
hope this helps

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43.
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(1 point) There were 23.7 million licensed drivers in California in 2009 and 22.76 million in 2004. Find a formula for the numbe

Salsk061 [2.6K]

Answer:

N(t) = 0.188t + 22.76

Step-by-step explanation:

Number of licensed drivers in 2004 = 22.76 million

Number of licensed drivers in 2009 = 23.7 million

Number of licensed drivers, N as a function of t since year 2004 ;

General form of a linear function :

y = mx + c

c = intercept ; m = slope

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22.76 = c

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We can find the slope :

y = m*5 + 22.76

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N(t) = 0.188t + 22.76

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2 years ago

The diameter of a particle of contamination (in micrometers) is modeled with the probability density function f(x)= 2/x^3 for x

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Answer:

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c) 0.018

d) 0.982

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We are given with the Probability density function f(x)= 2/x^3 where x > 1.

<em>Firstly we will calculate the general probability that of P(a < X < b) </em>

P(a < X < b) = $\int_{a}^{b} \frac{2}{x^{3}} dx$ = $2\int_{a}^{b} x^{-3} dx$

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= $2[ \frac{x^{-2} }{-2}]^{b}_a$ = $\frac{2}{-2} [ x^{-2} ]^{b}_a$

= $-1 [ b^{-2} - a^{-2} ]$ = $\frac{1}{a^{2} } - \frac{1}{b^{2} }$

a) Now P(X < 5) = P(1 < X < 5) {because x > 1 }

Comparing with general probability we get,

P(1 < X < 5) = $\frac{1}{1^{2} } - \frac{1}{5^{2} }$ = $1 - \frac{1}{25}$ = 0.96 .

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d) P(x < 6 or X > 10) = P(1 < X < 6) + P(10 < X < ∞)

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⇒ P(1 < X < x) = 0.75

⇒ $\frac{1}{1^{2} } - \frac{1}{x^{2} }$ = 0.75

⇒ $\frac{1} {x^{2} }$ = 1 - 0.75 = 0.25

⇒ $x^{2}$ = $\frac{1}{0.25}$ ⇒ $x^{2}$ = 4 ⇒ x = $2$

Therefore, value of x such that P(X < x) = 0.75 is 2.

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