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3 years ago

Please Help, I can't figure out this problem

Mathematics

1 answer:

Otrada [13]3 years ago

3 0

Answer:

<h2>The answer is 5 units</h2>

Step-by-step explanation:

The distance between two points can be found by using the formula

$d = \sqrt{ ({x1 - x2})^{2} + ({y1 - y2})^{2} } \\$

where

(x1 , y1) and (x2 , y2) are the points

From the question

The points are (9,-7) and (5, -4)

The distance between them is

$d = \sqrt{ ({9 - 5})^{2} + ({ - 7 + 4})^{2} } \\ = \sqrt{ {4}^{2} + ({ - 3})^{2} } \\ = \sqrt{16 + 9} \\ = \sqrt{25}$

We have the final answer as

<h3>5 units</h3>

Hope this helps you

You might be interested in

In Case Study 19.1, we learned that about 56% of American adults actually voted in the presidential election of 1992, whereas ab

Radda [10]

Answer:

a) Confidence interval for 68% confidence level

= (0.548, 0.572)

Confidence interval for 95% confidence level

= (0.536, 0.584)

Confidence interval for 99.99% confidence level = (0.523, 0.598)

b) The sample proportion of 0.61 is unusual as falls outside all of the range of intervals where the sample mean can found for all 3 confidence levels examined.

c) Standardized score for the reported percentage using a sample size of 400 = 2.02

Since, most of the variables in a normal distribution should fall within 2 standard deviations of the mean, a sample mean that corresponds to standard deviation of 2.02 from the population mean makes it seem very plausible that the people that participated in this sample weren't telling the truth. At least, the mathematics and myself, do not believe that they were telling the truth.

Step-by-step explanation:

The mean of this sample distribution is

Mean = μₓ = np = 0.61 × 1600 = 976

But the sample mean according to the population mean should have been

Sample mean = population mean = nP

= 0.56 × 1600 = 896.

To find the interval of values where the sample proportion should fall 68%, 95%, and almost all of the time, we obtain confidence interval for those confidence levels. Because, that's basically what the definition of confidence interval is; an interval where the true value can be obtained to a certain level.of confidence.

We will be doing the calculations in sample proportions,

We will find the confidence interval for confidence level of 68%, 95% and almost all of the time (99.7%).

Basically the empirical rule of 68-95-99.7 for standard deviations 1, 2 and 3 from the mean.

Confidence interval = (Sample mean) ± (Margin of error)

Sample Mean = population mean = 0.56

Margin of Error = (critical value) × (standard deviation of the distribution of sample means)

Standard deviation of the distribution of sample means = √[p(1-p)/n] = √[(0.56×0.44)/1600] = 0.0124

Critical value for 68% confidence interval

= 0.999 (from the z-tables)

Critical value for 95% confidence interval

= 1.960 (also from the z-tables)

Critical values for the 99.7% confidence interval = 3.000 (also from the z-tables)

Confidence interval for 68% confidence level

= 0.56 ± (0.999 × 0.0124)

= 0.56 ± 0.0124

= (0.5476, 0.5724)

Confidence interval for 95% confidence level

= 0.56 ± (1.960 × 0.0124)

= 0.56 ± 0.0243

= (0.5357, 0.5843)

Confidence interval for 99.7% confidence level

= 0.56 ± (3.000 × 0.0124)

= 0.56 ± 0.0372

= (0.5228, 0.5972)

b) Based on the obtained intervals for the range of intervals that can contain the sample mean for 3 different confidence levels, the sample proportion of 0.61 is unusual as it falls outside of all the range of intervals where the sample mean can found for all 3 confidence levels examined.

c) Now suppose that the sample had been of only 400 people. Compute a standardized score to correspond to the reported percentage of 61%. Comment on whether or not you believe that people in the sample could all have been telling the truth, based on your result.

The new standard deviation of the distribution of sample means for a sample size of 400

√[p(1-p)/n] = √[(0.56×0.44)/400] = 0.0248

The standardized score for any is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (0.61 - 0.56)/0.0248 = 2.02

Standardized score for the reported percentage using a sample size of 400 = 2.02

Hope this Helps!!!

7 0

3 years ago

Solve these recurrence relations together with the initial conditions given. a) an= an−1+6an−2 for n ≥ 2, a0= 3, a1= 6 b) an= 7a

8_murik_8 [283]

Answer:

a) 3/5·((-2)^n + 4·3^n)
b) 3·2^n - 5^n
c) 3·2^n + 4^n
d) 4 - 3 n
e) 2 + 3·(-1)^n
f) (-3)^n·(3 - 2n)
g) ((-2 - √19)^n·(-6 + √19) + (-2 + √19)^n·(6 + √19))/√19

Step-by-step explanation:

These homogeneous recurrence relations of degree 2 have one of two solutions. Problems a, b, c, e, g have one solution; problems d and f have a slightly different solution. The solution method is similar, up to a point.

If there is a solution of the form $a[n]=r^n$ , then it will satisfy ...

$r^n=c_1\cdot r^{n-1}+c_2\cdot r^{n-2}$

Rearranging and dividing by $r^{n-2}$ , we get the quadratic ...

$r^2-c_1r-c_2=0$

The quadratic formula tells us values of r that satisfy this are ...

$r=\dfrac{c_1\pm\sqrt{c_1^2+4c_2}}{2}$

We can call these values of r by the names r₁ and r₂.

Then, for some coefficients p and q, the solution to the recurrence relation is ...

$a[n]=pr_1^n+qr_2^n$

We can find p and q by solving the initial condition equations:

$\left[\begin{array}{cc}1&1\\r_1&r_2\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]$

These have the solution ...

$p=\dfrac{a[0]r_2-a[1]}{r_2-r_1}\\\\q=\dfrac{a[1]-a[0]r_1}{r_2-r_1}$

_____

Using these formulas on the first recurrence relation, we get ...

$c_1=1,\ c_2=6,\ a[0]=3,\ a[1]=6\\\\r_1=\dfrac{1+\sqrt{1^2+4\cdot 6}}{2}=3,\ r_2=\dfrac{1-\sqrt{1^2+4\cdot 6}}{2}=-2\\\\p=\dfrac{3(-2)-6}{-5}=\dfrac{12}{5},\ q=\dfrac{6-3(3)}{-5}=\dfrac{3}{5}\\\\a[n]=\dfrac{3}{5}(-2)^n+\dfrac{12}{5}3^n$

The rest of (b), (c), (e), (g) are solved in exactly the same way. A spreadsheet or graphing calculator can ease the process of finding the roots and coefficients for the given recurrence constants. (It's a matter of plugging in the numbers and doing the arithmetic.)

_____

For problems (d) and (f), the quadratic has one root with multiplicity 2. So, the formulas for p and q don't work and we must do something different. The generic solution in this case is ...

$a[n]=(p+qn)r^n$