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3 years ago

I’m confused, any help?

Mathematics

2 answers:

svetoff [14.1K]3 years ago

6 0

Answer:

Step-by-step explanation:

Trust me its D because to equal f(x)--->Inf., you need to have x equal a negative

irina [24]3 years ago

4 0

D hope this helps your welcome

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A football team was training for four hours. During the first hour, they practiced for 5/8 of an hour. During the second hour, t

Leviafan [203]

Answer: 2 19/24 hours was spent in practising.

Step-by-step explanation:

During the first hour, they practiced for 5/8 of an hour. During the second hour, they practiced for 2/3 of an hour. This means that the total time for which they practiced in the first 2 hours would be

5/8 + 2/3 = 31/24 hours

During the last two hours, they first practiced for 3/5 of an hour, took a 1/2 hour break and then practiced the rest of the time. This means that the rest of the time for which they practiced is

2 - (3/5 + 1/2) = 2 - 11/10 = 9/10

Therefore, the time they spent practicing in total would be

31/24 + 3/5 + 9/10 = 67/24 =

2 19/24 hours

7 0

3 years ago

Pls help me I reallyyyyy don't understand

mash [69]

Answer:

Step-by-step explanation:

range is the range from the lowest # through the biggest # in other words subtract the smaller number from the bigger number

8 0

3 years ago

On Texas Avenue between University Drive and George Bush Drive, accidents occur according to a Poisson process at a rate of thre

Zarrin [17]

Answer:

(a) The probability is 0.6514

(b) The probability is 0.7769

Step-by-step explanation:

If the number of accidents occur according to a poisson process, the probability that x accidents occurs on a given day is:

$P(x)=\frac{e^{-at}*(at)^{x} }{x!}$

Where a is the mean number of accidents per day and t is the number of days.

So, for part (a), a is equal to 3/7 and t is equal to 1 day, because there is a rate of 3 accidents every 7 days.

Then, the probability that a given day has no accidents is calculated as:

$P(x)=\frac{e^{-3/7}*(3/7)^{x}}{x!}$

$P(0)=\frac{e^{-3/7}*(3/7)^{0}}{0!}=0.6514$

On the other hand the probability that February has at least one accident with a personal injury is calculated as:

P(x≥1)=1 - P(0)

Where P(0) is calculated as:

$P(x)=\frac{e^{-at}*(at)^{x} }{x!}$

Where a is equivalent to (3/7)(1/8) because that is the mean number of accidents with personal injury per day, and t is equal to 28 because 4 weeks has 28 days, so:

$P(x)=\frac{e^{-(3/7)(1/8)(28)}*((3/7)(1/8)(28))^{x}}{x!}$

$P(0)=\frac{e^{-(3/7)(1/8)(28)}*((3/7)(1/8)(28))^{0}}{0!}=0.2231$

Finally, P(x≥1) is:

P(x≥1) = 1 - 0.2231 = 0.7769

3 0

4 years ago

Find the sum of the following series. Round to the nearest hundredth if necessary,

Aneli [31]

Answer:

$sum = \frac{a( {r}^{n - 1} )}{r - 1} \\ : but \: l = a( {r}^{n - 1} ) \\ 49152 = 3( {2}^{n - 1} ) \\ 16384 = {2}^{n - 1} \\ {2}^{n} = 32768 \\ {2}^{n} = {2}^{15} \\ n = 15 \\ \therefore \: sum = \frac{3(2 {}^{15 - 1}) }{15 - 1} \\ = \frac{49152}{14} \\ = 3510.9$