For this case, the first thing we must do is define variables:
x: unknown number (1)
y: unknown number (2)
We now write the equations that model the problem:
their sum is 6.1:
their difference is 1.6:
Solving the system we have:
We add both equations:
Then, we look for the value of y using any of the equations:
Answer:
The numbers are:
Answer:
what?
Step-by-step explanation:
Answer:
The percentage of students who scored below 620 is 93.32%.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean 
and standard deviation 
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this question, we have that:
Percentage of students who scored below 620:
This is the pvalue of Z when X = 620. So
has a pvalue of 0.9332
The percentage of students who scored below 620 is 93.32%.
If you throw a die, this is what you could get:
1, 2, 3, 4, 5 or 6.
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Odd Numbers: 1, 3, 5
Numbers less than Three: 1, 2
---------------
Possibility of getting 1, 2, 3 or 5:
4/6 which translates into 2/3.
---------------
Answer:
2/3
Answer:
(a) 169.1 m
Step-by-step explanation:
The diagram shows you the distance (x) will be shorter than 170 m, but almost that length. The only reasonable answer choice is ...
169.1 m
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The relevant trig relation is ...
Cos = Adjacent/Hypotenuse
The leg of the right triangle adjacent to the marked angle is x, and the hypotenuse is 170 m. Putting these values into the equation, you have ...
cos(6°) = x/(170 m)
x = (170 m)cos(6°) ≈ (170 m)(0.994522) ≈ 169.069 m
The horizontal distance covered is about 169.1 meters.
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<em>Additional comment</em>
Expressed as a percentage, the slope of this hill is tan(6°) ≈ 10.5%. It would be considered to be a pretty steep hill for driving.
