Answer:
a. The sampling distribution will be approximately normal.
d. The mean of the sampling distribution will be close to 52%
g. The standard deviation of the sampling distribution will be 0.0408
Step-by-step explanation:
For this problem the sample size is large enough (n>30), and then the sampling distribution 
would be approximately normal. The mean of the sampling distributions is given by 
The expected value for the sampling distribution would be 0.52 since 
And for the standard deviation we know that is given by:
So the correct answers on this case are:
a. The sampling distribution will be approximately normal.
d. The mean of the sampling distribution will be close to 52%
g. The standard deviation of the sampling distribution will be 0.0408
Let, the width = w
We know, A = 2(lw + wh +hl )
250 = 2(10w + 5w + 50)
250 = 2(15w + 50)
250 = 30w + 100
30w = 250 - 100
w = 150/30
w = 5
In short, Your Answer would be Option C) 5 cm
Hope this helps!
The LCM of the denominators is 120. In other words, the denominators can multiply to 120.
3 * 40 = 120
40 * 2 = 80
Tigers: P = 80/120
24 * 5 = 120
24 * 4 = 96
Redbirds: P = 96/120
8 * 15 = 120
3 * 15 = 45
Bulldogs P = 45/120
60 * 2 = 120
60 * 1 = 60
Titans P = 60/120
Answer
Tigers: P = 80/120
Redbirds: P = 96/120
Bulldogs: P = 45/120
Titans: P = 60/120
Answer:
B) 1
Step-by-step explanation:
The computation of the number is shown below:
Since the number 1 would be raised to any exponent so it always remains be 1
Like
= 1 × 1 × 1 × 1 × 1 × 1 × 1 × 1 × 1
= 1
Therefore the option B is correct
Answer:
Large Box = 18.5
Small Box = 15.75
Step-by-step explanation:
Let x = large box and y = small box
5x+6y=187
3x+2y=87
cancel out one equation
(5x+6y=187)•3=15x+18y=561
(3x+2y=87)•-5=15x+10y=435
15x+18y=561
-15x-10y=-435
–––––––––––
0+ 8y= 126
8y÷8 =126÷8
y=15.75
Substitute y into either original equation
3x+2(15.75)=87
3x+ 31.5=87
3x+ 31.5-31.5=87-31.5
3x=55.5
3x÷3=55.5÷3
x=18.5
