Answer:
2.18 repeating.
Step-by-step explanation:
So it will be 24/11 because the negatives cancel out each other. So the answer is 2.18 repeating.
Jay walks 2 1/4 miles which if you double it so the denominators are e same for both 2 1/4 and 1 1/8 then your answer will be 18/8 (Jay) and 9/8 (Reggie). Meaning that Reggie walks have the distance Jay walks.
Answer:
a. H0 : p ≤ 0.11 Ha : p >0.11 ( one tailed test )
d. z= 1.3322
Step-by-step explanation:
We formulate our hypothesis as
a. H0 : p ≤ 0.11 Ha : p >0.11 ( one tailed test )
According to the given conditions
p`= 31/225= 0.1378
np`= 225 > 5
n q` = n (1-p`) = 225 ( 1- 31/225)= 193.995> 5
p = 0.4 x= 31 and n 225
c. Using the test statistic
z= p`- p / √pq/n
d. Putting the values
z= 0.1378- 0.11/ √0.11*0.89/225
z= 0.1378- 0.11/ √0.0979/225
z= 0.1378- 0.11/ 0.02085
z= 1.3322
at 5% significance level the z- value is ± 1.645 for one tailed test
The calculated value falls in the critical region so we reject our null hypothesis H0 : p ≤ 0.11 and accept Ha : p >0.11 and conclude that the data indicates that the 11% of the world's population is left-handed.
The rejection region is attached.
The P- value is calculated by finding the corresponding value of the probability of z from the z - table and subtracting it from 1.
which appears to be 0.95 and subtracting from 1 gives 0.04998
Answer:
The length of each side of the base is
Step-by-step explanation:
we know that
The area of a square is equal to
where
b is the length side of a square
In this problem we have
so
substitute in the formula
Square root both sides
Answer:
4Joules
Step-by-step explanation:
According to Hooke's law which states that extension of an elastic material is directly proportional to the applied force provide that the elastic limit is not exceeded. Mathematically,
F = ke where
F is the applied force
K is the elastic constant
e is the extension
If a spring exerts a force of 6 N when stretched 3 m beyond its natural length, its elastic constant 'k'
can be gotten using k = f/e where
F = 6N, e = 3m
K = 6N/3m
K = 2N/m
Work done on an elastic string is calculated using 1/2ke².
If the spring is stretched 2 m beyond its natural length, the work done on the spring will be;
1/2× 2× (2)²
= 4Joules