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3 years ago

I need help QUICK! with this difficult question

Mathematics

1 answer:

likoan [24]3 years ago

4 0

Answer:

Step-by-step explanation:

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Day 2- Real World- Midpoint

nevsk [136]

Answer:

Point of the flower garden R ( 11, 44 )

Step-by-step explanation:

Let´s call P ( 3, 14 ) The center of the herb garden, and Q ( 7 , 29 ) the birdbath point

Differences in coordinates of these two points are:

x axis 7 - 3 = 4

y axis 29 - 14 = 15

Then the point Q ( x , y ) will be at: Q ( 11, 44)

x axis x = 11 because 7 + 4 = 11 means that the coordinate 7 is just halfway way between coordinates (x) 3 and 11 ( 4 above 3, and 4 below 11)

Reasoning the same way we conclude y = 44 in this case 29 is just halfway between coordinates (y) 14 and 44 ( 15 above 14 and 15 below 44)

We can probe getting distance d₁ between P and Q and d₂ distance between Q and R as:

d₁ = √ (7-3)² + ( 29-14)²

d₁ = √ 16 + 225

d₂ = √ (11 - 7)² + ( 44 - 29)²

d₂ = √ 16 + 225

3 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 ...

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

The initial condition equations are now ...

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

and the solutions for p and q are ...

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

Using these formulas on problem (d), we get ...

$c_1=2,\ c_2=-1,\ a[0]=4,\ a[1]=1\\\\r=\dfrac{2+\sqrt{2^2+4(-1)}}{2}=1\\\\p=4,\ q=\dfrac{1-4(1)}{1}=-3\\\\a[n]=4-3n$

And for problem (f), we get ...

$c_1=-6,\ c_2=-9,\ a[0]=3,\ a[1]=-3\\\\r=\dfrac{-6+\sqrt{6^2+4(-9)}}{2}=-3\\\\p=3,\ q=\dfrac{-3-3(-3)}{-3}=-2\\\\a[n]=(3-2n)(-3)^n$

_____

<em>Comment on problem g</em>

Yes, the bases of the exponential terms are conjugate irrational numbers. When the terms are evaluated, they do resolve to rational numbers.

6 0

3 years ago

Which probability distribution table corresponds with this frequency distribution table

nlexa [21]

Answer:

$\begin{array}{cccccc}x&1&2&3&4&5\\f&0.2&0.3&0.05&0.2&0.25\end{array}$

Step-by-step explanation:

First find the sum

$\sum f=4+6+1+4+5=20.$

Now, find the probabilities:

$Pr(x=1)=\dfrac{4}{20}=\dfrac{1}{5}=0.2;$
$Pr(x=2)=\dfrac{6}{20}=\dfrac{3}{10}=0.3;$
$Pr(x=3)=\dfrac{1}{20}=0.05;$
$Pr(x=4)=\dfrac{4}{20}=\dfrac{1}{5}=0.2;$
$Pr(x=5)=\dfrac{5}{20}=\dfrac{1}{4}=0.25.$

Hence, the frequency distribution table is

$\begin{array}{cccccc}x&1&2&3&4&5\\f&0.2&0.3&0.05&0.2&0.25\end{array}$

4 0

3 years ago

The U.S. Bureau of Labor Statistics reports that of persons who usually work full-time, the average number of hours worked per w

alukav5142 [94]

Answer:

The standard deviation of number of hours worked per week for these workers is 3.91.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by

$Z = \frac{X - \mu}{\sigma}$

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

In this problem we have that:

The average number of hours worked per week is 43.4, so $\mu = 43.4$ .

Suppose 12% of these workers work more than 48 hours. Based on this percentage, what is the standard deviation of number of hours worked per week for these workers.

This means that the Z score of $X = 48$ has a pvalue of 0.88. This is Z between 1.17 and 1.18. So we use $Z = 1.175$ .