Answer:
A) 34.13%
B) 15.87%
C) 95.44%
D) 97.72%
E) 49.87%
F) 0.13%
Step-by-step explanation:
To find the percent of scores that are between 90 and 100, we need to standardize 90 and 100 using the following equation:

Where m is the mean and s is the standard deviation. Then, 90 and 100 are equal to:

So, the percent of scores that are between 90 and 100 can be calculated using the normal standard table as:
P( 90 < x < 100) = P(-1 < z < 0) = P(z < 0) - P(z < -1)
= 0.5 - 0.1587 = 0.3413
It means that the PERCENT of scores that are between 90 and 100 is 34.13%
At the same way, we can calculated the percentages of B, C, D, E and F as:
B) Over 110

C) Between 80 and 120

D) less than 80

E) Between 70 and 100

F) More than 130

Answer:
i dont know if this is right but i think that would be the second one
Step-by-step explanation:
i dont know if this is right but i think that would be the second one
Answer:
1.) Unbalanced
Step-by-step explanation:
have a great day
Answer:
the error is in step 2.
0.25x was added to -0.75x with an incorrect result of +0.50x.
Step-by-step explanation:
The incorrect step is ...
Step 2: negative 1.50 = 0.50 x
This is the result of an attempt to add 0.25x to both sides of the equation. The correct step would be ...
-1.50 -0.25x = -0.75x . . . . . . . result from Step 1
-1.50 -0.25x +0.25x = -0.75x +0.25x . . . . . add 0.25x to both sides
-1.50 = -0.50x . . . . . . . . . . . . . the correct result from the addition (Step 2)
__
The correct result is ...
x = -1.50/-0.50 = 3 . . . . . Step 3
Answer:
21 ways
Step-by-step explanation:
number = 7 digit
5 digit no = 52115
to find out
How many different seven-digit numbers
solution
first we need to place the two missing 3s in the number 52115
we consider here two cases
case 1 the two 3's appear separated (like 532135 or 3521135)
case 2 the two 3's appear together (like 5332115 or 5211533)
Case 1 we can see that number type as _5_2_1_1_5_
place 3's placeholders show potential locations
( type a ) for 3's separated we will select 2 of 6 place and place 3 in every location so we do this 6C2 = (15) ways
and (type b): again use same step as _5_2_1_1_5_
here 3s together for criterion and we will select 1 of the 6 place and place both 3s here and there are 6 ways.
so that here will be 15+6=21 ways
If 3 and 3 are separate so 6C2 = 15 ways
If 3 and 3 are together so there = 6 ways
= 15 + 6 = 21 ways