May be there is an operator missing in the first function, h(x). I will solve this in two ways, 1) as if the h(x) = 5x and 2) as if h(x) = 5 + x
1) If h(x) = 5x and k(x) = 1/x
Then (k o h) (x) = k ( h(x) ) = k(5x) = 1/(5x)
2) If h(x) = 5 + x and k (x) = 1/x
Then (k o h)(x) =k ( h(x) ) = k (5+x) = 1 / [5 + x]
Answer:
Step-by-step explanation:
The given function is
To find k(-5), we put x=-5 into the given function to get;
Multiply out to obtain;
This is the same as;
This simplifies to
Opening brackets gives;
4-8v=-8v+4
Putting like terms together;
4-4=-8v+8v
0=0
There is no solution thus.
Answer:
the probability that the sample variance exceeds 3.10 is 0.02020 ( 2,02%)
Step-by-step explanation:
since the variance S² of the batch follows a normal distribution , then for a sample n of 20 distributions , then the random variable Z:
Z= S²*(n-1)/σ²
follows a χ² ( chi-squared) distribution with (n-1) degrees of freedom
since
S² > 3.10 , σ²= 1.75 , n= 20
thus
Z > 33.65
then from χ² distribution tables:
P(Z > 33.65) = 0.02020
therefore the probability that the sample variance exceeds 3.10 is 0.02020 ( 2,02%)
Since we are already given the amount of jumps from the first trial, and how much it should be increased by on each succeeding trial, we can already solve for the amount of jumps from the first through tenth trials. Starting from 5 and adding 3 each time, we get: 5 8 (11) 14 17 20 23 26 29 32, with 11 being the third trial.
Having been provided 2 different sigma notations, which I assume are choices to the question, we can substitute the initial value to see if it does match the result of the 3rd trial which we obtained by manual adding.
Let us try it below:
Sigma notation 1:
10
<span> Σ (2i + 3)
</span>i = 3
@ i = 3
2(3) + 3
12
The first sigma notation does not have the same result, so we move on to the next.
10
<span> Σ (3i + 2)
</span><span>i = 3
</span>
When i = 3; <span>3(3) + 2 = 11. (OK)
</span>
Since the 3rd trial is a match, we test it with the other values for the 4th through 10th trials.
When i = 4; <span>3(4) + 2 = 14. (OK)
</span>When i = 5; <span>3(5) + 2 = 17. (OK)
</span>When i = 6; <span>3(6) + 2 = 20. (OK)
</span>When i = 7; 3(7) + 2 = 23. (OK)
When i = 8; <span>3(8) + 2 = 26. (OK)
</span>When i = 9; <span>3(9) + 2 = 29. (OK)
</span>When i = 10; <span>3(10) + 2 = 32. (OK)
Adding the results from her 3rd through 10th trials: </span><span>11 + 14 + 17 + 20 + 23 + 26 + 29 + 32 = 172.
</span>
Therefore, the total jumps she had made from her third to tenth trips is 172.
