Mean is average
so we have 4 numbers
sum them then divide by 4
(32+31+37+44)=144
divide by 4
144/4=36
answer is C
Answer:
Step-by-step explanation:
Do substitution by first writing an equation 15c + 29a=175 when you get your answer either substitute it in your other equation which would be 21c+27a=177, if you found c(Children) in your first equation substitute it in 21 by multipying if you found a do the same thing.
Answer: (7, 20)
Concept:
There are three general ways to solve systems of equations:
- Elimination
- Substitution
- Graphing
Since the question has specific requirements, we are going to use <u>substitution </u>to solve the equations.
Solve:
<u>Given equations</u>
y = 3x - 1
2x + 6 = y
<u>Substitute the y value since both equations has isolated [y]</u>
2x + 6 = 3x - 1
<u>Add 1 on both sides</u>
2x + 6 + 1 = 3x - 1 + 1
2x + 7 = 3x
<u>Subtract 2x on both sides</u>
2x + 7 - 2x = 3x - 2x
<u>Find the value of y</u>
y = 3x - 1
y = 3(7) - 1
y = 21 - 1
Hope this helps!! :)
Please let me know if you have any questions
Answer:
Step-by-step explanation:
Hello!
X: Cholesterol level of a woman aged 30-39. (mg/dl)
This variable has an approximately normal distribution with mean μ= 190.14 mg/dl
1. You need to find the corresponding Z-value that corresponds to the top 9.3% of the distribution, i.e. is the value of the standard normal distribution that has above it 0.093 of the distribution and below it is 0.907, symbolically:
P(Z≥z₀)= 0.093
-*or*-
P(Z≤z₀)= 0.907
Since the Z-table shows accumulative probabilities P(Z<Z₁₋α) I'll work with the second expression:
P(Z≤z₀)= 0.907
Now all you have to do is look for the given probability in the body of the table and reach the margins to obtain the corresponding Z value. The first column gives you the integer and first decimal value and the first row gives you the second decimal value:
z₀= 1.323
2.
Using the Z value from 1., the mean Cholesterol level (μ= 190.14 mg/dl) and the Medical guideline that indicates that 9.3% of the women have levels above 240 mg/dl you can clear the standard deviation of the distribution from the Z-formula:
Z= (X- μ)/δ ~N(0;1)
Z= (X- μ)/δ
Z*δ= X- μ
δ=(X- μ)/Z
δ=(240-190.14)/1.323
δ= 37.687 ≅ 37.7 mg/dl
I hope it helps!
