Answer:
we cannot conclude hat the proportion of wives married less than two years who planned to have children is significantly higher than the proportion of wives married five years
Step-by-step explanation:
Given that in a study on the fertility of married women conducted by Martin O’Connell and Carolyn C. Rogers for the Census Bureau in 1979, two groups of childless wives aged 25 to 29 were selected at random, and each was asked if she eventually planned to have a child. One group was selected from among wives married less than two years and the other from among wives married five years.
Let X be the group married less than 2 years and Y less than 5 years
X Y Total
Sample size 300 300 600
Favouring 240 288 528
p 0.8 0.96 0.88

p difference = -0.16
Std error for difference = 
Test statistic = p difference/std error=-6.03
p value <0.000001
Since p is less than alpha 0.05 we cannot conclude hat the proportion of wives married less than two years who planned to have children is significantly higher than the proportion of wives married five years
Alright, lets get started.
If Matthew wants to complete packages at an average rate of at least 39 packages per hour.
And they worked 4 hrs only due to picnic, yesterday, it means they have to make
packages.
But they made only 112 packages means they are short of
packages.
Suppose they are working today t hrs, and his department will complete 43 packages per hour today.
It means they are going to make 43 t packages today.
This 43 t packages includes those 44 too , which they are short of yesterday due to picnic.
So, average will be
(39 average given in question)
Cross multiplying

Adding 44 in both sides


Subtracting 39 t in both sides


Dividing 4 in both sides
t = 11 hrs
Hence they have to woth 11 hrs today : Answer
Hope it will help :)
I = p*r*t
$3000=p*.06*5=?
Multiply and you would have yor answer(rounded to nearsest cent)!
He soled 21 and 1/2 of almonds at the fair because 25 is = to 24 and 2/2 so that - 3 and 1/2 is = to 21 and 1/2
Step-by-step explanation:
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