Ements

Solve the inequality

krok68 [10]

4b <= - 3
b <= -3/4

answer is A
b <= -3/4

7 0

3 years ago

What is the sum of A and 8???

dolphi86 [110]

Answer:

11

Step-by-step explanation:

a+b÷3-2=11 jsiwidiifiwjdjwjxjdjjw

6 0

3 years ago

Farmer Ed has 2,500 meters of fencing,

Juliette [100K]

Answer:

Because it is a rectangle, the area is expressed as A = xy, or length times width.

Because it is next to the river, he only needs to fence three sides, so F = x + 2y.

Knowing the amount of fencing available is 7500m, we get:

7500 = x + 2y solve for x

x = 7500 - 2y substitute into the area equation

A = (7500 - 2y)y distribute

A = -2y2 +7500y

You can see that this is a parabola which opens down, meaning that the point of maximum area will be at the vertex, y = -b/2a = -7500/[2(-2)] = 1875

x = 7500 - 2(1875) = 3750

A = 3750(1875) = 7,031,250 m2

Step-by-step explanation:

4 0

4 years ago

The first quartile of a data set is 41, and the third quartile is 57. Which of these values in the data set is an outlier?

ikadub [295]

The answer is C............

3 0

4 years ago

Over 500 million tweets are sent per day (digital Marketing ramblings website, December 15, 2014). Assume that the number of twe

Aleksandr [31]

Answer:

(a) The probability that Bob receives no tweets during his lunch hour is 0.0002.

(b) The probability that Bob receives at least 4 tweets during his lunch hour is 0.9190.

(c) The expected number of tweets Bob receives during the first 30 minutes of his lunch hour is 3.5.

(d) The probability that Bob receives no tweets during the first 30 minutes of his lunch hour is 0.0302.

Step-by-step explanation:

Let X = number of tweets.

The random variable X follows a Poisson distribution with parameter λ = 7.

The probability mass function of a Poisson distribution is:

$P(X=x)=\frac{e^{-7}(7)^{x}}{x!};\ x=0, 1, 2, 3...$

(a)

Compute the probability that Bob receives no tweets during his lunch hour as follows:

$P(X=0)=\frac{e^{-7}(7)^{0}}{0!}\\=\frac{0.000192\times1}{1} \\=0.000192\\\approx0.0002$

Thus, the probability that Bob receives no tweets during his lunch hour is 0.0002.

(b)

Compute the probability that Bob receives at least 4 tweets during his lunch hour as follows:

P (X ≥ 4) = 1 - P (X < 4)

= 1 - P (X = 0) - P (X = 1) - P (X = 2) - P (X = 3)

$=1-\frac{e^{-7}(7)^{0}}{0!}-\frac{e^{-7}(7)^{1}}{1!}-\frac{e^{-7}(7)^{2}}{2!} - \frac{e^{-7}(7)^{3}}{3!}\\=1-0.0002-0.0064-0.0223-0.0521\\=0.9190$

Thus, the probability that Bob receives at least 4 tweets during his lunch hour is 0.9190.

(c)

The average number of tweets in 60 minutes is 7.

Then the average number of tweets in 1 minute is, $\frac{7}{60}$ .

Hence, the average number of tweets during 30 minutes is, $\frac{7}{60}\times30=3.5$

Thus, the expected number of tweets Bob receives during the first 30 minutes of his lunch hour is 3.5.

(d)

Compute the probability that Bob receives no tweets during the first 30 minutes of his lunch hour as follows:

$P(No\ tweets) = \frac{e^{-3.5}(3.5)^{0}}{0!}=\frac{0.0302\times1}{1} =0.0302$

Thus, the probability that Bob receives no tweets during the first 30 minutes of his lunch hour is 0.0302.

3 0

4 years ago