Answer:
6.18% of the class has an exam score of A- or higher.
Step-by-step explanation:
When the distribution is normal, we use 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:
What percentage of the class has an exam score of A- or higher (defined as at least 90)?
This is 1 subtracted by the pvalue of Z when X = 90. So
has a pvalue of 0.9382
1 - 0.9382 = 0.0618
6.18% of the class has an exam score of A- or higher.
Answer:
12 + 2x
Step-by-step explanation:
<em>simplifying</em><em> </em><em>expression</em><em>:</em>
2(10) + 2(x-4)
2 × 10 = 20
2 × x = 2x
2 × -4 = -8
20 + 2x - 8
20 - 8 = 12
12 + 2x
<em>if you want to simplify it</em><em> </em><em>more</em><em>:</em>
(12/2) + (2x/2)
2(6 + x)
<em>if you want to solve for x:</em>
12 + 2x = 0
(12 - 12) + 2x = (0 - 12)
2x/2 = -12/2
x = -6
hope this helps you!
The numbers are: "3" and "9" .
_______________________________________________________
Explanation:
_______________________________________________________
Let "x" represent one of the two (2) numbers.
Let "y" represent the other one of the two (2) numbers.
x = 2y + 3 ;
x + y = 12 .
__________________________
Method 1)
__________________________
x = 12 <span>− y ;
Plug this into "x" for "2y + 3 = x" ;
</span>→ 2y + 3 = 12 <span>− y ;
</span>
Add "y" to each side of the equation; & subtract "3" from each side of the equation ;
→ 2y + 3 + y − 3 = 12 − y + y <span>− 3 ;
</span>
to get: 3y = 9 ;
Divide each side of the equation by "3" ;
to isolate "y" on one side of the equation; & to solve for "y" ;
3y / 3 = 9 / 3 ;
y = 3 .
____________________________
Now: x = 12 − y ; Plug in "3" for "y" ; to solve for "x" ;
→ x = 12 − 3 = 9
____________________________
So; x = 9, y = 3 .
____________________________
Method 2)
____________________________
When we have:
____________________________
x = 2y + 3 ;
x + y = 12 .
____________________________
→ y = 12 − x ;
_____________________________
Substitute "(12−x)" for "y" in the equation:
" x = 2y + 3 " ;
→ x = 2(12 − x) + 3 ;
_____________________________________
Note the "distributive property of multiplication" :
_____________________________________
a(b + c) = ab + ac ;
a(b − c) = ab − ac ;
_____________________________________
As such:
_____________________________________
→ 2(12 − x) = 2(12) − 2(x) = 24 − 2x ;
_____________________________________
So; rewrite:
_____________________________________
x = 2(12 − x) + 3 ;
as:
_____________________________________
→ x = 24 − 2x + 3 ;
→ x = 27 − 2x ;
Add "2x" to each side of the equation:
→ x + 2x = 27 − 2x + 2x ;
→ 3x = 27 ;
Divide each side of the equation by "3" ;
to isolate "x" on one side of the equation; & to solve for "x" ;
3x / 3 = 27 / 3 ;
x = 9 .
___________________________
Note: "y = 12 − x" ; Substitute "9" for "x" ; to solve for "y" ;
→ y = 12 − 9 = 3 ;
→ y = 3 .
__________________________
So, x = 9 ; and y = 3.
____________________________
The numbers are: "3" and "9" .
_____________________________
To check our answers:
Let us plug these numbers into the original equations;
to see if the equations hold true ; (i.e. when, "x = 9" ; and "y = 3"
_____________________________
→ x + y = 12 ;
→ 9 + 3 =? 12 ? Yes!
_____________________________
→ x = 2y + 3 ;
→ 9 =? 2(3) + 3 ?? ;
→ 9 =? 6 + 3 ? Yes!!
_____________________________
Answer:
If you went to the party, then you would have received a gift.
Step-by-step explanation:
?
