Answer:
yes they are
Step-by-step explanation:
because the metric system is the same
Answer:
P ( 5 < X < 10 ) = 1
Step-by-step explanation:
Given:-
- Sample size n = 49
- The sample mean u = 8.0 mins
- The sample standard deviation s = 1.3 mins
Find:-
Find the probability that the average time waiting in line for these customers is between 5 and 10 minutes.
Solution:-
- We will assume that the random variable follows a normal distribution with, then its given that the sample also exhibits normality. The population distribution can be expressed as:
X ~ N ( u , s /√n )
Where
s /√n = 1.3 / √49 = 0.2143
- The required probability is P ( 5 < X < 10 ) minutes. The standardized values are:
P ( 5 < X < 10 ) = P ( (5 - 8) / 0.2143 < Z < (10-8) / 0.2143 )
= P ( -14.93 < Z < 8.4 )
- Using standard Z-table we have:
P ( 5 < X < 10 ) = P ( -14.93 < Z < 8.4 ) = 1
727.29 + 248.50 − x ≥ 500;
x ≥ $475.79
727.29 + 248.50 − x ≤ 500;
x ≤ $475.79
727.29 − 248.50 + x ≥ 500
x ≥ $21.21
727.29 – 248.50 - x ≤ 500
x ≤ $21.21
Answer:
I don't know soorry. lalalala
Answer:
x=-2
Step-by-step explanation:
To solve this equation, we can use PEMDAS or Order of Operations.
Parenthesis
Exponents
Multiplication>Division
Addition>Subtraction
Using the various properties can also help make the equation easier.
First, solve for parenthesis using the distributive property.
Our equation is now : 3/4x+3=1/4x+2
Now, subtract 2 on both sides, to cancel out the positive 2 on the right.
3/4x+1=1/4x
Now subtract 3/4x from both sides.
1=-2/4x
Finally, to isolate x, divide both sides by -2/4
1/-2/4=-2
x=-2
Hope this helps!
