Answer:
center = 98.6, variability = 0.08
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a random variable X, with mean 
and standard deviation 
, the sample means with size n of at least 30 can be approximated to a normal distribution with mean and standard deviation 
The center is the mean.
So 
The standard deviation of the sample of 50 adults is the variability, so
So the correct answer is:
center = 98.6, variability = 0.08
Answer:
Step-by-step explanation:
2x + 3.5x + 15 = -1.5
5.5x + 15 = -1.5
5.5x = -16.5
x = -3
y = 3.5(-3) + 15
= -10.5 + 15
= 4.5
(-3, 4.5)
Let t represent Todd's age now.
.. 4(t -3) -(t -3) = 81 . . . . . . 3 years ago, their differnce in ages was 81.
.. 3t -9 = 81
.. t = (81 +9)/3 = 30
Todd is 30 now.
_____
You can also work this by considering "ratio units." 3 years ago, the ratio of their ages was 4:1, a difference of 3. That difference corresponds to 81 years, so each "ratio unit" represents 81/3 = 27 years. Todd's age then was 1 ratio unit, 27 years. Now, Todd's age is 30.
1. You have that:
- The<span> lengths of the bases are (6x-1) units and 3 units.
- The midsegment has a length of (5x-3) units.
2. To solve this exercise, you must apply the formula for calculate the length of the midsegment of a trapezoid, which is shown below:
Midsegment=Base1+Base2/2
As you can see, the midsegment is half the sum of the bases of the trapezoid.
3. When you substitute the values, you obtain:
(5x-3)=[(6x-1)+3]/2
4. Now, you can solve the problem by clearing the "x":
</span>
(5x-3)=[(6x-1)+3]/2
2(5x-3)=6x-1+3
10x-6=6x+2
10x-6x=2+6
4x=8
x=8/4
x=2
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.
