Answer:
The probability that sample mean differ the true greater than 2.1 will be 2.8070 %
Step-by-step explanation:
Given:
Sample mean =46 dollars
standard deviation=8
n=53
To Find :
Probability that sample mean would differ from true mean by greater than 2.1
Solution;
<em>This sample distribution mean problem,</em>
so for that
calculate Z- value
Z=(sample mean - true mean)/(standard deviation/Sqrt(n))
Z=-2.1/(8/Sqrt(53))
Z=-2.1*Sqrt(53)/8
Z=-1.91102
Now for P(X≥2.1)=P(Z≥-1.91102)
Using Z-table,
For Z=-1.91
P(X>2.1)=0.02807
Answer:
29°
Step-by-step explanation:
∠1 ≅ ∠8
3x+25 = 4x -17
42 = x . . . . . . . . . . add 17-3x
Now, we can find the measure of ∠1:
m∠1 = 3(42) +25 = 151
Angle 2 is the supplement to this, so has measure ...
m∠2 = 180° -151° = 29°
Interquartile range is the range is the number in the dead center, you have to divide the number line into 2 sections. The middle of everything and the middle of both section is the interquartile range
Hope this helps!
Answer:
1) 1/17
2) 1/111
3) 1/15
They're all fractions
Hope that helped, just divide the result with the number available
Example :
1÷17 =1/17
Hi!
<h3>Put in the values. </h3>
[z+8(5-2)]^2/4
<h3>Now use PEMDAS to solve. </h3>
- Parentheses - Subtraction
[z+8*3]^2/4
- Parentheses - Multiplication
[z+24]^2/4
z+576/4
<u>z+144</u>
<h2>When you evaluate [a+8 (b-2)]^2÷4, you get z + 144</h2>
Hope this helps! :)
-Peredhel
