Answer:
z= 3.63
z for significance level = 0.05 is ± 1.645
Step-by-step explanation:
Here p = 42% = 0.42
n= 500
We formulate our null and alternative hypotheses as
H0: p= 0.42 against Ha : p> 0.42 One tailed test
From this we can find q which is equal to 1-p= 1-0.42 = 0.58
Taking p`= 0.5
Now using the z test
z= p`- p/ √p(1-p)/n
Putting the values
z= 0.5- 0.42/ √0.42*0.58/500
z= 0.5- 0.42/ 0.0220
z= 3.63
For one tailed test the value of z for significance level = 0.05 is ± 1.645
Since the calculated value does not fall in the critical region we reject our null hypothesis and accept the alternative hypothesis that more than 42% people owned cats.
<span> 16501
(10^23) • 16501/5000
</span><span> 2.1 10 = 2•5
</span>
<span>(10)^23 = (2•5)^23 = 2^23 • 5^23</span>
<span> 16501
(2^23•5^23) • 16501/5000
</span>
Answer:
y = -3x - 1
Use the methods on your other questions.
Yes bc 6 groups and a bout 4 are in it ...........
4/24= 1/6
(1/6)*84= 14 had nuts
20/24= 5/6
(5/6)* 84 = 70 no nuts
