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A school survey reported the percent of students in Zoe’s class who ride their bikes to school. The percent is shown below. This

ZanzabumX [31]

Answer:

the answwer is D

Step-by-step explanation:

YOUR WELCOME!!!!

6 0

3 years ago

When you round a number to the nearest hundred you get 13,600 and when you round it to the nearest thousand you get 14,00.What w

Viefleur [7K]

The original number would be 13,555 .

6 0

3 years ago

7 copies of the sum of 8 fifths and 4

marta [7]

Answer:

39.2

Step-by-step explanation:

We need to find the solution of 7 copies of sum of 8 fifths and 4.

Now, 8 fifths = $\frac{8}{5}$

Sum of 8 fifths and 4 = = $\frac{28}{5}$

Now, 7 copies of $\frac{28}{5}$ means that we meed to multiply $\frac{28}{5}$ 7 times

7 copies = $\frac{196}{5}$

7 copies = 39.2

Hence, the solution is 39.2.

8 0

3 years ago

Read 2 more answers

Which of the following is the value of x? A. 2 B.11 C. 20 D. 29

Lunna [17]

Answer:

c

Step-by-step explanation:

180= 45+ (5x+35)

-45 -45

135 = 5x + 35

-35 -35

100 = 5x

/5 /5

x=20

5 0

3 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