The total amount is 12 ounces therefore you can use that as your denominator.
4 ounces out of it are vanilla and the other 8 are chocolate.
4 out of 12 is vanilla . This can also be written as 4/12.
4/12 can be your first fraction.
The second one can be a simplified version of 4/12.
The simplified version would be 1/3.
Your two fractions would be 4/12 and 1/3.
Hope This Helps You!
Good Luck Studying :)
Answer:
Last part 24x
Step-by-step explanation:
I understand some of this, like the last question. if each class had 24 pupil, and were trying to find out how many are in each class, it would be 24x (x stands for the amount of classes)
You have not given a true amount of pupils so I can not give a 100% correct answer for the first part.
Answer:
H0: μm − μw = 0
against the claim
Ha: μm − μw ≠ 0
Since the calculated value of z= 0.6177 does not lie in the critical region the null hypothesis is accepted that men and women have equal success in challenging calls.
Step-by-step explanation:
1) Let the null and alternate hypothesis be
H0: μm − μw = 0
against the claim
Ha: μm − μw ≠ 0
2) The significance level is set at 0.05
3) The critical region is z > + 1.96 and z< -1.96
4) The test statistic
Z= p1-p2/ sqrt [pcqc( 1/n1+ 1/n2)]
Here p1= 411/ 1390= 0.2956 and p2= 213/753=0.2829
pc = 411+ 213/1390+753
pc=624/2143
pc= 0.2912
qc= 1-pc= 1-0.2912=0.7088
5) Calculations
Z= p1-p2/ sqrt [pcqc( 1/n1+ 1/n2)]
z= 0.2956-0.2829/√ 0.2912*0.7088( 1/1390+ 1/753)
z= 0.0127/ √0.2064 (0.00204)
z= 0.0127/0.02056
z= 0.6177
6) Conclusion
Since the calculated value of z= 0.6177 does not lie in the critical region the null hypothesis is accepted that men and women have equal success in challenging calls.
If the function is:
y=a cos( bx+c), or y=a sin(x+c)
then the period is T= 2π/b and if T=2:
2=2π/b
b=2π/2=π
Answer: B) y=3 cos πx
8-5+10 divided by 5*6 + 2 = 17
