Answer:
Step-by-step explanation:
Step 1:-
We have c1(t) = e^ t i + (sin(t))j + t³k
and c2(t) = e^−t i + (cos(t))j − 6t³k.
By adding c1(t) and c2(t):
c1(t)+c2(t) = e^ t i + (sin(t))j + t³k + e^−t i + (cos(t))j − 6t³k
Now, employing the derivative formula:
Next, differentiate with respect to 't'
By factoring out i, j, and k terms, we arrive at:
Answer:
The P-value ranges between 2.5% and 5% according to the t-table.
Step-by-step explanation:
A random sample of 16 students from a large university showed an average age of 25 years with a standard deviation of 2 years.
Let = true average age of all students at the university.
So, the Null Hypothesis, :
24 years {indicating the average age is less than or equal to 24 years}
Alternate Hypothesis, :
> 24 years {indicating the average age is significantly greater than 24 years}
Here we employ the One-sample t-test statistics as the population's standard deviation is unknown;
T.S. = ~
where, = sample average age = 25 years
s = sample standard deviation = 2 years
n = sample size = 16
This gives us the test statistics = ~
= 2
The value of the t-test statistics is 2.
Moreover, the P-value of the test-statistics can be found as follows;
P-value = P( > 2) = 0.034 {as per the t-table}
Thus, the P-value lies between 2.5% and 5% based on the t-table.
Answer:
The number of standard deviations above the mean is
Step-by-step explanation:
The question indicates that:
The average weight of the corn ears from each farm is
The standard deviation for the corn ears from Iowa is
The standard deviation for the corn ears from Ohio is
A randomly chosen ear of corn from Iowa weighs x = 1.39 pounds
The standardized score is z = 1.645
The weight of a randomly chosen ear of corn from Ohio measures
In general, the standardized score of corn weight from Iowa can be mathematically defined as:
=>
=>
=>
Conversely, the standardized score of corn weight from Ohio is expressed as:
=>
=>
=>
A positive value indicates this quantity represents the number of standard deviations above the mean.