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Airida [17]
2 years ago
11

Emily waited in line for the StarCraft video game for 3 hours. If the line was only moving 2 1⁄4 feet per hour, how far did Emil

y move in the line?
Mathematics
2 answers:
BabaBlast [244]2 years ago
5 0

Answer:

6 3/4 feet

Step-by-step explanation:

Harrizon [31]2 years ago
3 0

Answer:

6 3/4 feet

Step-by-step explanation:

2 1/4 x 3 = 6 3/4

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Shade the model to show the decimal 0.542 using 100 grid and 10 grid
kherson [118]

0.542=0.500 + 0.040+0.002

          = \frac{5}{10} + \frac{4}{100} +\frac{2}{1000}

         =5 tenth + 4 hundredths + two thousandths

One tenth = \frac{1}{10}

One Hundredth= \frac{1}{100}

One thousandth= \frac{1}{1000}


8 0
3 years ago
Read 2 more answers
the ratio of the circumferences of two circles is 2:3. If the large circle has a radius of 39 cm, what is the radiusof the small
Orlov [11]

C1/C2 =  2/3

 2pi r1 / 2pi r2 = r1/r2 = 2/3

r1/39 = 2/3

r1 = 39*2/3 = 26

So, your answer is 26cm

8 0
3 years ago
Read 2 more answers
1. A report from the Secretary of Health and Human Services stated that 70% of single-vehicle traffic fatalities that occur at n
Nuetrik [128]

Using the binomial distribution, it is found that there is a 0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

For each fatality, there are only two possible outcomes, either it involved an intoxicated driver, or it did not. The probability of a fatality involving an intoxicated driver is independent of any other fatality, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

In this problem:

  • 70% of fatalities involve an intoxicated driver, hence p = 0.7.
  • A sample of 15 fatalities is taken, hence n = 15.

The probability is:

P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)

Hence

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 10) = C_{15,10}.(0.7)^{10}.(0.3)^{5} = 0.2061

P(X = 11) = C_{15,11}.(0.7)^{11}.(0.3)^{4} = 0.2186

P(X = 12) = C_{15,12}.(0.7)^{12}.(0.3)^{3} = 0.1700

P(X = 13) = C_{15,13}.(0.7)^{13}.(0.3)^{2} = 0.0916

P(X = 14) = C_{15,14}.(0.7)^{14}.(0.3)^{1} = 0.0305

P(X = 15) = C_{15,15}.(0.7)^{15}.(0.3)^{0} = 0.0047

Then:

P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) = 0.2061 + 0.2186 + 0.1700 + 0.0916 + 0.0305 + 0.0047 = 0.7215

0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

A similar problem is given at brainly.com/question/24863377

5 0
2 years ago
Y = -2x + 2 y = 5x + 9​
WITCHER [35]
X=-1 y=4 hope it helps
8 0
2 years ago
What is 0.08 divided by 1.44
hammer [34]
0.08 divided by 1.44 is 0.055555556
6 0
3 years ago
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