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Zina [86]
3 years ago
5

Jared drove a distance of 165 miles to billings last weekend, which used 11 gallons of gas. How many gallons of gas will he need

to drive to denver, 735 miles away?
Mathematics
1 answer:
blondinia [14]3 years ago
3 0

Answer:

8550

Step-by-step explanation:

1. you need to figure out how many time 11 goes into 165 which is 15 then you need to get the 15 and multiply it by 570.

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Step-by-step explanation:

a- no u don't need to know

b- 38.7m-3.9m=34.8m

c-3.48+3.9= 38.7

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Answer:

B. The population is declining at a rate of 11.5%

Step-by-step explanation:

if your graph the equation you will clearly see that it is declining and if you don't know how to graph the equation then go to the demos calculator, its an app & website, super easy to use. most people would assume the answer would be one of the 88.5% just because it's in the original equation but the .885 in the equation doesn't represent the rate of change there for the answer is 11.5%.

5 0
3 years ago
businessText message users receive or send an average of 62.7 text messages per day. How many text messages does a text message
KiRa [710]

Answer:

(a) The probability that a text message user receives or sends three messages per hour is 0.2180.

(b) The probability that a text message user receives or sends more than three messages per hour is 0.2667.

Step-by-step explanation:

Let <em>X</em> = number of text messages receive or send in an hour.

The random variable <em>X</em> follows a Poisson distribution with parameter <em>λ</em>.

It is provided that users receive or send 62.7 text messages in 24 hours.

Then the average number of text messages received or sent in an hour is: \lambda=\frac{62.7}{24}= 2.6125.

The probability of a random variable can be computed using the formula:

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

(a)

Compute the probability that a text message user receives or sends three messages per hour as follows:

P(X=3)=\frac{e^{-2.6125}(2.6125)^{3}}{3!} =0.21798\approx0.2180

Thus, the probability that a text message user receives or sends three messages per hour is 0.2180.

(b)

Compute the probability that a text message user receives or sends more than three messages per hour as follows:

P (X > 3) = 1 - P (X ≤ 3)

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

             =1-\frac{e^{-2.6125}(2.6125)^{0}}{0!}-\frac{e^{-2.6125}(2.6125)^{1}}{1!}-\frac{e^{-2.6125}(2.6125)^{2}}{2!}-\frac{e^{-2.6125}(2.6125)^{3}}{3!}\\=1-0.0734-0.1916-0.2503-0.2180\\=0.2667

Thus, the probability that a text message user receives or sends more than three messages per hour is 0.2667.

6 0
3 years ago
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