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Digiron [165]
3 years ago
15

Fraction (12 4/9 + 1 2/9) +3 5/9

Mathematics
1 answer:
sveticcg [70]3 years ago
6 0
12 4/9 + 1 2/9=13 6/9+3 5/9=16 11/9 that answer is farther simpled 17 2/9
so the final answer is 17 2/9
You might be interested in
In the final quarter of 2011, a company received 32% of the total profit for all mobile phone manufacturers. If the total profit
Sav [38]

The total profit received for all mobile phone manufacturers for all carriers was 1.54 billion

Step-by-step explanation:

Given,

Profit received = 32%

Total profit received = $4.8 billion

Amount of profit received for mobiles = 32% of total profit

Amount received = \frac{32}{100}*4.8

Amount received = \frac{153.6}{100}

Amount received = 1.536 billion

Rounding off

Amount received = 1.54 billion

The total profit received for all mobile phone manufacturers for all carriers was 1.54 billion

Keywords: percentage, division

Learn more about percentage at:

  • brainly.com/question/10940255
  • brainly.com/question/10941043

#LearnwithBrainly

3 0
3 years ago
Suppose a geyser has a mean time between irruption’s of 75 minutes. If the interval of time between the eruption is normally dis
lesya [120]

Answer:

(a) The probability that a randomly selected Time interval between irruption is longer than 84 minutes is 0.3264.

(b) The probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is 0.0526.

(c) The probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is 0.0222.

(d) The probability decreases because the variability in the sample mean decreases as we increase the sample size

(e) The population mean may be larger than 75 minutes between irruption.

Step-by-step explanation:

We are given that a geyser has a mean time between irruption of 75 minutes. Also, the interval of time between the eruption is normally distributed with a standard deviation of 20 minutes.

(a) Let X = <u><em>the interval of time between the eruption</em></u>

So, X ~ Normal(\mu=75, \sigma^{2} =20)

The z-score probability distribution for the normal distribution is given by;

                            Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean time between irruption = 75 minutes

           \sigma = standard deviation = 20 minutes

Now, the probability that a randomly selected Time interval between irruption is longer than 84 minutes is given by = P(X > 84 min)

 

    P(X > 84 min) = P( \frac{X-\mu}{\sigma} > \frac{84-75}{20} ) = P(Z > 0.45) = 1 - P(Z \leq 0.45)

                                                        = 1 - 0.6736 = <u>0.3264</u>

The above probability is calculated by looking at the value of x = 0.45 in the z table which has an area of 0.6736.

(b) Let \bar X = <u><em>sample time intervals between the eruption</em></u>

The z-score probability distribution for the sample mean is given by;

                            Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean time between irruption = 75 minutes

           \sigma = standard deviation = 20 minutes

           n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is given by = P(\bar X > 84 min)

 

    P(\bar X > 84 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{84-75}{\frac{20}{\sqrt{13} } } ) = P(Z > 1.62) = 1 - P(Z \leq 1.62)

                                                        = 1 - 0.9474 = <u>0.0526</u>

The above probability is calculated by looking at the value of x = 1.62 in the z table which has an area of 0.9474.

(c) Let \bar X = <u><em>sample time intervals between the eruption</em></u>

The z-score probability distribution for the sample mean is given by;

                            Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean time between irruption = 75 minutes

           \sigma = standard deviation = 20 minutes

           n = sample of time intervals = 20

Now, the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is given by = P(\bar X > 84 min)

 

    P(\bar X > 84 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{84-75}{\frac{20}{\sqrt{20} } } ) = P(Z > 2.01) = 1 - P(Z \leq 2.01)

                                                        = 1 - 0.9778 = <u>0.0222</u>

The above probability is calculated by looking at the value of x = 2.01 in the z table which has an area of 0.9778.

(d) When increasing the sample size, the probability decreases because the variability in the sample mean decreases as we increase the sample size which we can clearly see in part (b) and (c) of the question.

(e) Since it is clear that the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is very slow(less than 5%0 which means that this is an unusual event. So, we can conclude that the population mean may be larger than 75 minutes between irruption.

8 0
2 years ago
(6c^2-c+1)-(-4c+2c^2+8c)
Natalka [10]
6c^2 - c + 1 + 4c - 2c^2 - 8c
= 4c^2 - 5c + 1
= (4c - 1)(c - 1)
3 0
3 years ago
A dice is rolled 5 times. How many possible rolls are there?
ch4aika [34]

Answer:

7776

Step-by-step explanation:

Yes.

7 0
2 years ago
Whats 9 +10 will give brainlest
shtirl [24]

Answer:

19

Step-by-step explanation:

"You stupid"

"No Im not"

"Whats 9+10"

"21"

"You stupid"

4 0
3 years ago
Read 2 more answers
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