Jill gets paid every two weeks her last paycheck was 1503.50 if she earns 24.25 per hour how many hours did she work

Now, as we want to know when are the per capita out-of-pocket expenses for health care predicted to be $1400, and this total is our variable y, then

$x=\frac{1400-543}{39.6}\approx 21.6$

Finally, we know that x is the number of years past the year 2000, so the answer is that during the year 2021, the per capita out-of-pocket expenses for health care are predicted to be $1400.

3 0

4 years ago

For every integer k from 1 to 10, inclusive the "k"th term of a certain sequence is given by (−1)(k+1)∗(12k). If T is the sum of

Katena32 [7]

Answer:

Option D. is the correct option.

Step-by-step explanation:

In this question expression that represents the kth term of a certain sequence is not written properly.

The expression is $(-1)^{k+1}(\frac{1}{2^{k}})$ .

We have to find the sum of first 10 terms of the infinite sequence represented by the expression given as $(-1)^{k+1}(\frac{1}{2^{k}})$ .

where k is from 1 to 10.

By the given expression sequence will be $\frac{1}{2},\frac{(-1)}{4},\frac{1}{8}.......$

In this sequence first term "a" = $\frac{1}{2}$

and common ratio in each successive term to the previous term is 'r' = $\frac{\frac{(-1)}{4}}{\frac{1}{2} }$

r = $-\frac{1}{2}$

Since the sequence is infinite and the formula to calculate the sum is represented by

$S=\frac{a}{1-r}$ [Here r is less than 1]

$S=\frac{\frac{1}{2} }{1+\frac{1}{2}}$

$S=\frac{\frac{1}{2}}{\frac{3}{2} }$

S = $\frac{1}{3}$

Now we are sure that the sum of infinite terms is $\frac{1}{3}$ .

Therefore, sum of 10 terms will not exceed $\frac{1}{3}$

Now sum of first two terms = $\frac{1}{2}-\frac{1}{4}=\frac{1}{4}$

Now we are sure that sum of first 10 terms lie between $\frac{1}{4}$ and $\frac{1}{3}$

Since $\frac{1}{2}>\frac{1}{3}$

Therefore, Sum of first 10 terms will lie between $\frac{1}{4}$ and $\frac{1}{2}$ .

Option D will be the answer.

3 0

3 years ago