Answer:
Each day, you record the high temperature* (you don't necessarily need to say that it is for a science project). On Day 1, it was 4* less than Day 10**. It hit 62* F on Day 10. What was the high temperature on Day 1?
Because we know that on Day 1, the temperature was 4* less than on Day 10, and the high temperature on Day 10 was 62*, the high temperature on Day 1 was therefore 58*. Because 62 - 4 = 58.
Step-by-step explanation:
*You don't necessarily need to say that it was for a science project.
**The information for Day 4 is irrelevant too because we are not given the temperature for Day 4, therefore it does not help us find the answer.
Answer:
16
Step-by-step explanation:
To solve, we can use an equation finding the average of the siblings'ages.
45/117
Both numbers can be divide by 9
5/13
Remember, you can do anything to an equation as long as you do it to both sides
4+x-17=22
add like terms
x+4-17=22
x-13=22
add 13 to both sides
x+13-13=22+13
x+0=35
x=35
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.
