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Bond [772]
3 years ago
5

Below are the results of tossing a number cube 7times. Find the experimental probability of tossing an even number. 6 4 3 2 5 3

3
Mathematics
2 answers:
Daniel [21]3 years ago
7 0
3/7 or 3:7 because you tossed the dice 7 times and got an even number 3 of those times.
den301095 [7]3 years ago
4 0

Answer:

Probability of tossing an even number is \frac{3}{7}

Step-by-step explanation:

Given : The results of tossing a number cube 7 times is 6 4 3 2 5 3 3 .

To find : The experimental probability of tossing an even number.      

Solution :  

\text{Probability}=\frac{\text{Favorable outcome}}{\text{Total number of outcome}}

In the given result of tossing a cube 7 times we get,

6 4 3 2 5 3 3

Tossing an even number - 6,4,2

Favorable outcome (tossing an even number) = 3

Total number of outcome = 7

Probability of tossing an even number is

\text{Probability}=\frac{3}{7}

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THERE WAS A HALF CLASS OF WATER IN A FRACTION HOW MUCH WATER WOULD 3 PEOPLE GET?
kkurt [141]

Answer:

0.16

Step-by-step explanation:

you divide 1/2 by 3 to 0.16666666 but it would recuce to 0.16

6 0
4 years ago
Kate sold 12 of her paintings and purchased 7 more. if kate has 21 paintings now, how many paintings did she have to begin with?
kicyunya [14]
First, let x or any variable represent the number of paintings Kate initially has. After selling 12 of it, her paintings now becomes only x - 12. Then, she purchased 7 more making the number of her paintings become x - 12 + 7 = x - 5. This number tantamount to 21. 

                                           x - 5 = 21      ; x = 26

Thus, Kate initially had 26 paintings. 
7 0
3 years ago
-7(y-1)= 21<br><br> Show how you answered the question step by step.
Stolb23 [73]

Step-by-step explanation:

-7(y-1)= 21

-7y+7=21

-7y=21-7

-7y=14

divide by-7

y=-2

6 0
4 years ago
Read 2 more answers
The mean number of words per minute (WPM) read by sixth graders is 8888 with a standard deviation of 1414 WPM. If 137137 sixth g
Bingel [31]

Noticing that there is a pattern of repetition in the question (the numbers are repeated twice), we are assuming that the mean number of words per minute is 88, the standard deviation is of 14 WPM, as well as the number of sixth graders is 137, and that there is a need to estimate the probability that the sample mean would be greater than 89.87.

Answer:

"The probability that the sample mean would be greater than 89.87 WPM" is about \\ P(z>1.56) = 0.0594.

Step-by-step explanation:

This is a problem of the <em>distribution of sample means</em>. Roughly speaking, we have the probability distribution of samples obtained from the same population. Each sample mean is an estimation of the population mean, and we know that this distribution behaves <em>normally</em> for samples sizes equal or greater than 30 \\ n \geq 30. Mathematically

\\ \overline{X} \sim N(\mu, \frac{\sigma}{\sqrt{n}}) [1]

In words, the latter distribution has a mean that equals the population mean, and a standard deviation that also equals the population standard deviation divided by the square root of the sample size.

Moreover, we know that the variable Z follows a <em>normal standard distribution</em>, i.e., a normal distribution that has a population mean \\ \mu = 0 and a population standard deviation \\ \sigma = 1.

\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}} [2]

From the question, we know that

  • The population mean is \\ \mu = 88 WPM
  • The population standard deviation is \\ \sigma = 14 WPM

We also know the size of the sample for this case: \\ n = 137 sixth graders.

We need to estimate the probability that a sample mean being greater than \\ \overline{X} = 89.87 WPM in the <em>distribution of sample means</em>. We can use the formula [2] to find this question.

The probability that the sample mean would be greater than 89.87 WPM

\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}}

\\ Z = \frac{89.87 - 88}{\frac{14}{\sqrt{137}}}

\\ Z = \frac{1.87}{\frac{14}{\sqrt{137}}}

\\ Z = 1.5634 \approx 1.56

This is a <em>standardized value </em> and it tells us that the sample with mean 89.87 is 1.56<em> standard deviations</em> <em>above</em> the mean of the sampling distribution.

We can consult the probability of P(z<1.56) in any <em>cumulative</em> <em>standard normal table</em> available in Statistics books or on the Internet. Of course, this probability is the same that \\ P(\overline{X} < 89.87). Then

\\ P(z

However, we are looking for P(z>1.56), which is the <em>complement probability</em> of the previous probability. Therefore

\\ P(z>1.56) = 1 - P(z

\\ P(z>1.56) = P(\overline{X}>89.87) = 0.0594

Thus, "The probability that the sample mean would be greater than 89.87 WPM" is about \\ P(z>1.56) = 0.0594.

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ludmilkaskok [199]

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