Let us assume the unknown number in the question to be = x
Then
160% * x = 32
(160/100) * x = 32
16x/10 = 32
16x = 32 * 10
16x = 320
x = 320/16
= 20
So from the above deduction we can easily conclude that the unknown number in the question is 20. Be very careful about the calculation and the cross multiplication part of the problem. Other than that there is nothing complicated in the question.
The mean would be 5
Reason for this is that the mean is when all data is summed up then divided by how many data points there are
So 4+4+7=15
Then divide by 3 would equal 5
I hope this helps!
<h2>Answer:
y = - ¹/₂ x + 5
</h2>
<h3>Step-by-step explanation:
</h3>
<u>Find the slope of the perpendicular line</u>
When two lines are perpendicular, the product of their slopes is -1. This means that the slopes are negative-reciprocals of each other.
⇒ if the slope of this line = 2 (y = 2x + 2)
then the slope of the perpendicular line (m) = - ¹/₂
<u>Determine the equation</u>
We can now use the point-slope form (y - y₁) = m(x - x₁)) to write the equation for this line:
⇒ y - 3 = - ¹/₂ (x - 4)
We can also write the equation in the slope-intercept form by making y the subject of the equation and expanding the bracket to simplify:
since y - 3 = - ¹/₂ (x - 4)
y = - ¹/₂ x + 5 (in slope-intercept form)
Answer:
as posted: 0.18182
more likely: 2.00
Step-by-step explanation:
z = (x -μ)/σ
... = (90 -76)/77 = 14/77 = 2/11
z ≈ 0.18182
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We know that cutting and pasting often results in numbers being unintentionally repeated, so your 77 may be intended to be 7. In that case, the z-score is
... z = (90 -76)/7 = 14/7
z = 2
So there are two different GPA’s and as being an athlete I need to know what both are. Your weighted GPA is going to be the GPA that counts heavy on honors and AP classes so for instance if you get an A on an Honors test it will affect your grade and GPA more than a reg class sitting at a 1.88 or a 1.97 isn’t going to be good enough to get into most schools and if that is by the end of your freshman year it isn’t really going to move by the end of your senior year. Hope this helps!!!