Which graph represents the solution of the system?

Alexis buys groceries using her credit card her card has an APR of 19.99% and she must pay at least 5% of the balance at the end

avanturin [10]

Answer:

The answer is that she would pay $65.56 in finance charges at the end of the month.

Step-by-step explanation:

Given: APR = 19.99%

Carry Over Balance: $398.97

The APR or Annual Percentage Rate, is calculated daily. You will need to get the daily periodic rate, or DPR, so divide the APR by 365:

19.99% = .1999

.1999 / 365 = .005477 (This is the Approximate DPR, rounded up to .005477)

To get the finance charge, multiply the average daily balance by the DPR and then by 30 days:

398.97 * .005477 * 30 = $65.56 finance charge for this carry over balance, at the end of the month. This assumes that the balance is the average daily balance.

Hope this helps!! Have a great day!

7 0

3 years ago

Solve this simultaneous equation by method of elimination

lawyer [7]

Answer:

y

=

−

2

x

3

+

1

3

y

=

−

3

x

5

+

1

5

Step-by-step explanation:

7 0

3 years ago

Question is in the picture<br>please help

LiRa [457]

When given the graph of a function, the domain would include all the points that there is a graph. The strategy is to find what <em>is not</em> included.

What we are looking for are points of discontinuity. Think of it as when you remove your pencil from the paper.

From left to right, the graph stops at x = -3. So anything less than -3 is in the domain. Next, the graph starts up again at x =-1 after an asymptote (the vertical dashed lines). This piece goes to x = 4. So our domain is from -1 to 4.

Lastly, there's a jump from 4 to 5 and the graph goes on again. After 5, we take all the stuff more than it. So x > 5 is in the domain.

So x < -3, - 1 < x < 4, and x > 5 appears to be our domain. However, end points needed to be checked to see if we include them or not. Again we go left to right.

At x = -3 there is a filled (or closed) circle and that means we include -3.

At x = -1 there is an asymptote. Asymptotes are things you get close to but don't get to. (Think of it as the "I'm Not Touching" game you play on car trips.) So we exclude -1.

At x = 4 there is an unfilled (or open) circle and that means we exclude 4.

At x = 5 there is a filled circle so we include 5.

Now we refine our domain for the endpoints.

x ≤ -3, -1 < x < 4, x ≥ 5 is our domain.

The problem gives us intervals, and we gave it in inequalities. When we include an endpoint we use brackets - [ and } and when we exclude and endpoint we use parentheses - ( and ). Let's go back to x ≤ -3. Anything less works, and -3 is included (closed circle). That interval is (-∞, -3]. Next is the piece between -1 and 4. Since both are excluded, (-1,4) is our interval. We include 5 to write x ≥5 as the interval [5,∞).

Put the bolded ones all together and use the union, ∪, symbol to connect them, since something on the graph could be in any piece.

Our domain is (-∞, -3] ∪(-1,4) ∪ [5,∞).

5 0

3 years ago

What is an expression that uses partial products to multiply 8 x 64

Ivahew [28]

Ur answer to this question would be 512.

7 0

2 years ago

A study was conducted and two types of engines, A and B, were compared. Fifty experiments were performed using engine A and 75 u

USPshnik [31]

Answer:

a) -6 mpg.

b) 2.77 mpg

c) The 95% confidence interval for the difference of population mean gas mileages for engines A and B and interpret the results, in mpg, is (-8.77, -3.23).

Step-by-step explanation:

To solve this question, we need to understand the central limit theorem, and subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

Gas mileage A: Mean 36, standard deviation 6, sample of 50:

So

$\mu_A = 36, s_A = \frac{6}{\sqrt{50}} = 0.8485$

Gas mileage B: Mean 42, standard deviation 8, sample of 50:

So

$\mu_B = 42, s_B = \frac{8}{\sqrt{50}} = 1.1314$

Distribution of the difference:

Mean:

$\mu = \mu_A - \mu_B = 36 - 42 = -6$

Standard error:

$s = \sqrt{s_A^2+s_B^2} = \sqrt{0.8485^2+1.1314^2} = 1.4142$

A. Find the point estimate.

This is the difference of means, that is, -6 mpg.

B. Find the margin of error

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.025 = 0.975$ , so Z = 1.96.

Now, find the margin of error M as such

$M = zs = 1.96*1.4142 = 2.77$

The margin of error is of 2.77 mpg

C. Construct the 95% confidence interval for the difference of population mean gas mileages for engines A and B and interpret the results(5 pts)

The lower end of the interval is the sample mean subtracted by M. So it is -6 - 2.77 = -8.77 mpg

The upper end of the interval is the sample mean added to M. So it is -6 + 2.77 = -3.23 mpg

The 95% confidence interval for the difference of population mean gas mileages for engines A and B and interpret the results, in mpg, is (-8.77, -3.23).

8 0

2 years ago