Where the heck did x come from!
Answer:
Mean = 30516.67
Standard deviation, s = 3996.55
P(x < 27000) = 0.0011518
Step-by-step explanation:
Given the data:
28500 35500 32600 36000 34000 25700 27500 29000 24600 31500 34500 26800
Mean, xbar = Σx / n = 366200 /12 = 30516.67
Standard deviation, s = [√Σ(x - xbar) / n-1]
Using calculator, s = 3996.55
The ZSCORE = (x - mean) / s/√n
Zscore = (27000 - 30516.67) / (3996.55/√12)
Zscore = - 3516.67 / 1153.7046
Zscore = - 3.048
P(x < 27000) = P(Z < - 3.049) = 0.0011518
Answer: 7/8
What you have to do here is to find the highest number which can be divided by both. Or you have to divide both the numerator and denominator until it can not be divided by any number anymore. You can do this by finding the highest common factor (<u>HCF)</u> of both the number, 35 and 40.
Highest Common Factor: HCF of two or more numbers is the greatest factor that divides the numbers. For example, 2 is the HCF of 4 and 6.
So, the highest number which divides 35 and 40, is 5. Now 35 divided by 5 is 7 and 40 divided by 5 is 8. The final answer is 7/8 and there is no more numbers which can divide both of them by a specific number.
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hope it helped
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Answer:
Riley did better than Darby because One-half < Three-fifths.
Step-by-step explanation:
The statement that explains which girl did better would be that
Riley did better than Darby because One-half < Three-fifths.
There are a total of 60 minutes in an hour, If we multiply each of these fractions by 60 minutes it would give us the exact number of minutes that each girl took to finish the race.
Riley: 
* 60 = 30 minutes
Darby: 
* 60 = 36 minutes
Therefore, we can see that it took Riley 6 minutes less than Darby to finish the race meaning that she had the better result.
In order to move a number that is multiplied or divided by the variable<span>, you must do the inverse to both sides. That means, if the number is multiplied, you must divide it by both sides. If it's divided, you must </span>multiply<span> it by both side. steps so that you can apply them to more complex </span>problems<span>.</span>
