The health care Jimmy’s employer offers is a fee-for-service plan in which Jimmy and his family must pay for $12,500 in health related services in a calendar year before the insurance company begins to pay.  Jimmy and his family have several health related issues that demand $685.00 each month.  If one of Jimmy’s children were to become seriously injured at the end of the year, how much more would Jimmy and his family need to pay before the insurance company begins  to assume responsibility?a.
$984.58
b.
$4,280.00
c.
$11,815.00
d.
$14,417.92

Answer:

76 teenagers went to movies in the past month.

Step-by-step explanation:

The number of teens who went to movie out of every 5 teens  = 4

Total teens surveyed = 95

Let, out of 95 teenagers , the teens who went to movie = m

So, the ratio  of people who went : who were surveyed = 4: 5

Now,  BY THE LAW OF RATIO AND PROPORTIONALITY

or,

So, m = 76

⇒ 76 teenagers went to movies in the past month.

Answer:

b

Step-by-step explanation:

Step-by-step explanation:

To make the function f(x) = {sin(1/x), x ≠ 0; k, x = 0} continuous at x = 0, we need to find the value of k that ensures the limit of f(x) as x approaches 0 exists and is equal to k.

First, let's find the limit of sin(1/x) as x approaches 0:

lim(x -> 0) sin(1/x)

This limit does not exist because sin(1/x) oscillates wildly as x gets closer to 0. Therefore, in order for the function to be continuous at x = 0, we need to choose k such that it compensates for the oscillations of sin(1/x) as x approaches 0.

A suitable choice for k is 0 because the limit of sin(1/x) as x approaches 0 is undefined, and setting k = 0 ensures that f(x) becomes a continuous function at x = 0.

So, the correct choice is:

d. None (k = 0)

Final answer:

The value of k that would make the function f(x) = sin(1/x) when x ≠0 and f(x) = k when x=0 continuous at x=0 doesn't exist. This is because the limit of sin(1/x) as x approaches 0 is undefined, hence the function cannot be made continuous at x = 0 for any value of k.

Explanation:

To find the value of k that makes the function continuous at x=0, we can apply the definition of continuity, which states that a function, f(x), is continuous at a certain point, x0, if three conditions are met:

• the function is defined at x0
• the limit as x approaches x0 of f(x) exists
• the limit as x approaches x0 of f(x) is equal to f(x0)

In the case of the function f(x) = sin(1/x), the value for x = 0 is undefined, but we've been given that f(0) = k. To make the function continuous at x = 0, the value of k should ideally be equal to the limit of sin(1/x) as x approaches 0.

However, as x approaches 0, sin(1/x) oscillates between -1 and 1, making the limit non-existent. Because the limit does not exist, the function is not continuous at x=0 no matter the chosen value of k. Therefore, the correct answer is (d) None.

Learn more about Limits and Continuity here:

brainly.com/question/32625617

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Answer:

A.

Step-by-step explanation:

We want to find the Z-score of if the population mean is ,and the population standard deviation is .

We use the formula:

We substitute the values to obtain:

The correct answer is A.

Answer:

  25.  (x, y) = (5, 11)

  26.  (x, y) = (-1, 1)

Step-by-step explanation:

Both equations are of the form y=( ), so you can set the expressions for y equal to each other. Or, you can subtract the equation with the smaller y-coefficient from the other one.

25.

  x +6 = y = 2x +1 . . . . . equate expressions for y

  5 = x . . . . . . . . . . . subtract x+1

  y = 5+6 = 11 . . . . . using the first equation to find y

  (x, y) = (5, 11)

__

26.

  (y) -(y) = (3x +4) -(x+2) . . . . subtract the first equation from the second

  0 = 2x +2 . . . . . . . . . . . . . . simplify

  0 = x + 1 . . . . . . . . . . . . . . . . divide by the x-coefficient

  x = -1 . . . . . . . . . . . . . . subtract the constant

  y = -1 +2 = 1 . . . . . . . . . use the first equation to find y

  (x, y) = (-1, 1)

_____

Of course, when we say "subtract ..." or "divide ..." we mean that you should do the same operation to both sides of the equation. That way the equal sign remains valid. You can always use an expression or variable in place of its equal (this is the substitution property of equality).

The expression (x+1) that we subtract in problem 25 is the smaller x-term plus the constant on the opposite side of the equal sign. That way, we eliminate both the unwanted x-term and the unwanted constant. You can do these operations one at a time (and you were probably taught to do it that way). That is, subtract x; subtract 1.

For 26, the method of solution that puts both the variable and the constant on the same side of the equation and 0 on the other side has certain advantages. Subtracting one side of the equation from both sides (to make an expression equal to zero) will always work, regardless of the expressions involved. After simplification, you can divide by the coefficient of the variable to get the form x+constant=0, and the answer is always x = -constant. These simple instructions require no judgment. You may find it easier to choose to subtract the side with the smaller coefficient, so the result has a positive coefficient. That's not necessary, but it can reduce anxiety and errors.