If you’re asking what -11 times -6 is it’s going to be 66!
Answer:
An estimate
Step-by-step explanation:
It can not be simplified because the numerator and denominator can not be divided between the same number, so that each results an integer number.
Answer: 6 snack bags.
Step-by-step explanation:
For this exercise it is important to remember the Greatest Common Divisor (GCD). This is defined as the largest positive intener by which you can divided evenly "n" numbers.
In this case, you know that Jane has 18 juice boxes, 24 bags of peanuts and 36 apples. Therefore, the steps are the following:
1. You need to descompose 18, 24 and 36 into their prime factors, as below:
2. Now you must choose the common ones with the smallest exponent and multiply them.
Then, you get:
Therefore, the greatest number of snack bags that Jane can make is: 6 snack bags.
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.
