Let f(x) = 2x2 - 5x + 9. Evaluate the polynomial substituting 3 for x. Find the remainder. 11 12 13 14

A car rental company's standard charge includes an initial fee plus an additional fee for each mile driven. The standard charge

Radda [10]

4 days later. ugh haha sorry. answer in the first ten min on time my butt

7 0

3 years ago

Liz brought chocolate and vanilla cupcakes to school for her birthday. 18 students picked a vanilla cupcake and 42 students pick

romanna [79]

First, find the total number of students. There are 60 students total in the class. Now, do 18/60. This is .3. From here, multiply that by 100. 30% of students chose vanilla cupcakes. Hope this helped!

3 0

3 years ago

Solve the system by graphing. y = x+2 2x-y=-1

N76 [4]

Answer:

The intersection is at (1,3) and the answer is in the graph. (see image below)

Step-by-step explanation:

5 0

2 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

Two cars leave the same location traveling in opposite directions. One car leaves at 3:00 p.m. traveling at an average rate of 5

Misha Larkins [42]

Answer:

1

Step-by-step explanation:

8 0

3 years ago