A teacher surveys 30 students on the number of books they read over the summer. The data

In a West Texas school district the school year began on August 1 and lasted until May 31. On August 1 a Soft Drink company inst

WINSTONCH [101]

Answer:

$95.78

Step-by-step explanation:

f(t) = 300t / (2t² + 8)

t = 0 corresponds to the beginning of August. t = 1 corresponds to the end of August. t = 2 corresponds to the end of September. So on and so forth. So the second semester is from t = 5 to t = 10.

$T₂ = ∫₅¹⁰ 300t / (2t² + 8) dt

$T₂ = ∫₅¹⁰ 150t / (t² + 4) dt

$T₂ = 75 ∫₅¹⁰ 2t / (t² + 4) dt

$T₂ = 75 ln(t² + 4) |₅¹⁰

$T₂ = 75 ln(104) − 75 ln(29)

$T₂ ≈ 95.78

3 0

2 years ago

Two functions are defined as shown. f(x) =-1/2 x – 2 g(x) = –1 Which graph shows the input value for which f(x) = g(x)?

Natalka [10]

Answer:

See attachment.

Step-by-step explanation:

The given functions are:

$f(x) = - \frac{1}{2}x - 2$

and g(x)=-1

To find the x-value of the point of intersection of the two functions, we equate the two functions and solve for x.

$- \frac{1}{2}x - 2 = -1$

$x + 4 = 2$

$x = 2 - 4 = - 2$

The graph that shows the input value for which f(x)=g(x) is the graph which shows the point of intersection of f(x) and g(x) to be at x=-2.

4 0

2 years ago

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The number of flaws in a fiber optic cable follows a Poisson distribution. It is known that the mean number of flaws in 50m of c

boyakko [2]

Answer:

(a) The probability of exactly three flaws in 150 m of cable is 0.21246

(b) The probability of at least two flaws in 100m of cable is 0.69155

(c) The probability of exactly one flaw in the first 50 m of cable, and exactly one flaw in the second 50 m of cable is 0.13063

Step-by-step explanation:

A random variable X has a Poisson distribution and it is referred to as Poisson random variable if and only if its probability distribution is given by

$p(x;\lambda)=\frac{\lambda e^{-\lambda}}{x!}$ for x = 0, 1, 2, ...

where $\lambda$ , the mean number of successes.

(a) To find the probability of exactly three flaws in 150 m of cable, we first need to find the mean number of flaws in 150 m, we know that the mean number of flaws in 50 m of cable is 1.2, so the mean number of flaws in 150 m of cable is $1.2 \cdot 3 =3.6$

The probability of exactly three flaws in 150 m of cable is

$P(X=3)=p(3;3.6)=\frac{3.6^3e^{-3.6}}{3!} \approx 0.21246$

(b) The probability of at least two flaws in 100m of cable is,

we know that the mean number of flaws in 50 m of cable is 1.2, so the mean number of flaws in 100 m of cable is $1.2 \cdot 2 =2.4$

$P(X\geq 2)=1-P(X$

$P(X\geq 2)=1-p(0;2.4)-p(1;2.4)\\\\P(X\geq 2)=1-\frac{2.4^0e^{-2.4}}{0!}-\frac{2.4^1e^{-2.4}}{1!}\\\\P(X\geq 2)\approx 0.69155$

(c) The probability of exactly one flaw in the first 50 m of cable, and exactly one flaw in the second 50 m of cable is

$P(X=1)=p(1;1.2)=\frac{1.2^1e^{-1.2}}{1!}\\P(X=1)\approx 0.36143$

The occurrence of flaws in the first and second 50 m of cable are independent events. Therefore the probability of exactly one flaw in the first 50 m and exactly one flaw in the second 50 m is

$(0.36143)(0.36143) = 0.13063$

4 0

2 years ago

Stu hiked a trail at an average rate of 3 miles per hour. He ran back on the same trail at an average rate of 5 miles per hour.

Andrews [41]

3t = 5(3-t)
3t = 15 - 5t
3t + 5t = 15
8t = 15
t = 15/8
t = 1 7/8

7/8 * 60 mins = 52.50 minutes

t = 1 hour and 52.5 minutes

It took Stu 1 hour and 52.5 minutes to hike the trail one way.

3 0

2 years ago

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Which pair of variables shows a clear causal link?

xxTIMURxx [149]

The number of lights switched on in a room and the room's brightness.

8 0

2 years ago

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