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mr Goodwill [35]
3 years ago
14

Write the number in word form and in standard form 300,000,000+40,000,000+7,000,000+300,000+10,000+6,000+20+9

Mathematics
1 answer:
Citrus2011 [14]3 years ago
4 0
Three hundread and forty-seven million three hundread and sixteen thousand and twenty-nine....347,316,029
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If a ball is thrown in the air with a velocity 48 ft/s, its height in feet t seconds later is given by y = 48t − 16t2. Find the
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1. {26, 69, 30, 27, 19, 54, 27}<br><br> Mean: <br> , Median: <br> , Mode: <br> , Range:
den301095 [7]

Answer:

Mean:36

Median:27

Mode:27

Range:50

Step-by-step explanation:

Mean: 19+26+27+27+30+53+69/7=36

Median: The middle number in the sequence when arranged in ascending order which is 27

Mode: The number that appeared more than others which is 27

Range: This is the difference between the largest number and the lowest number in the sequence which is 50.

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2 years ago
ben plays basketball every 6 times in 3 weeks. At this rate, how many times does he play basketball in 10 weeks
Ratling [72]

Answer:

18

Step-by-step explanation:

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6 0
3 years ago
Read 2 more answers
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
2 years ago
Find the volume of the figure below.
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When you plug in r &h you get pi(6^2)(19.6/3) =
pi(36)(6.53)=738.9
8 0
2 years ago
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