Ask question

Ask question

All categories

2 years ago

13

To rationalize the denominator of StartFraction 5 minus StartRoot 7 EndRoot Over 9 minus StartRoot 14 EndRoot EndFraction , you

should multiply the expression by which fraction?

2 answers:

bekas [8.4K]2 years ago

8 0

Answer:

c

Step-by-step explanation:

just took the edgenuity quiz

hope this helps!

TEA [102]2 years ago

7 0

Answer:

the answer is the 3rd one

Step-by-step explanation:

i jusst took the test on edge

You might be interested in

From generation to generation, the mean age when smokers first start to smoke varies. However, the standard deviation of that ag

Jlenok [28]

Answer:

If we compare the p value and the significance level given for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we to reject the null hypothesis, and the the actual mean starting age of smokers is significantly lower than 19.

Step-by-step explanation:

1) Data given and notation

$\bar X=18.1$ represent the mean age when smokers first start to smoke varies

$s=1.3$ represent the standard deviation for the sample

$\sigma=2.1$ represent the population standard deviation

$n=37$ sample size

$\mu_o =5.7$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to determine if the mean starting age is at least 19, the system of hypothesis would be:

Null hypothesis: $\mu\geq 19$

Alternative hypothesis: $\mu < 19$

We know the population deviation, so for this case is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{18.1-19}{\frac{2.1}{\sqrt{37}}}=-2.607$

Calculate the P-value

Since is a one-side lower test the p value would be:

$p_v =P(z$

Conclusion

If we compare the p value and the significance level given for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we to reject the null hypothesis, and the the actual mean starting age of smokers is significantly lower than 19.

4 0

2 years ago

The donaldson furniture company produces three types of rocking chairs: the children's model, the standard model, and the exe

kirza4 [7]

<span>the question is 3 variable equation solution the three equation will be 5x+4y+7z=152 3x+2y+5z=92 2x+2y+4z=74 solving by elimination we get x=11,y=12,z=7</span>

5 0

1 year ago

Mr. Peterson needs topsoil for his garden. His rectangular garden is 78 in long and 48 in wide. A bag of topsoil covers an area

nata0808 [166]

Answer:

3744 inches squared, and 8 bags

Step-by-step explanation:

Ⓗⓘ ⓣⓗⓔⓡⓔ

˜”*°•.˜”*°• Area: •°*”˜.•°*”˜

Well, the formula for area is L*W

L=78 inches

W=48 inches

78*48=3744 inches squared

˜”*°•.˜”*°• Number of bags: •°*”˜.•°*”˜

3744/500=7.448

So he would need to buy 8 bags

(っ◔◡◔)っ ♥ Hope this helped! Have a great day! :) ♥

4 0

1 year ago

What is the approximate length of arc s on the circle below? Use 3.14 for Pi. Round your answer to the nearest tenth.

slega [8]

Answer:

The approximate length of arc s is 14.1 inches

Step-by-step explanation:

<u><em>The picture of the question in the attached figure</em></u>

step 1

Find the circumference of the circle

The formula to calculate the circumference is equal to

$C=2\pi r$

we have

$r=6\ in\\\pi=3.14$

substitute

$C=2(3.14)(6)\\C=37.68\ in$

step 2

Find the approximate length of arc s

we know that

The circumference of a circle subtends a central angle of 360 degrees

so

using proportion

Find the arc length s for a central angle of 135 degrees

$\frac{37.68}{360}=\frac{x}{135}\\\\x=37.68(135)/360\\\\x= 14.1\ in$

4 0

1 year ago

Read 2 more answers

Power series of y''+x^2y'-xy=0

Ray Of Light [21]

Assuming we're looking for a power series solution centered around $x=0$ , take

$y=\displaystyle\sum_{n\ge0}a_nx^n$
$y'=\displaystyle\sum_{n\ge1}na_nx^{n-1}$
$y''=\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}$

Substituting into the ODE yields

$\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}+\sum_{n\ge1}na_nx^{n+1}-\sum_{n\ge0}a_nx^{n+1}=0$

The first series starts with a constant term; the second series starts at $x^2$ ; the last starts at $x^1$ . So, extract the first two terms from the first series, and the first term from the last series so that each new series starts with a $x^2$ term. We have

$\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}=2a_2+6a_3x+\sum_{n\ge4}n(n-1)a_nx^{n-2}$

$\displaystyle\sum_{n\ge0}a_nx^{n+1}=a_0x+\sum_{n\ge1}a_nx^{n+1}$

Re-index the first sum to have it start at $n=1$ (to match the the other two sums):

$\displaystyle\sum_{n\ge4}n(n-1)a_nx^{n-2}=\sum_{n\ge1}(n+3)(n+2)a_{n+3}x^{n+1}$

So now the ODE is

$\displaystyle\left(2a_2+6a_3x+\sum_{n\ge1}(n+3)(n+2)a_{n+3}x^{n+1}\right)+\sum_{n\ge1}na_nx^{n+1}-\left(a_0x+\sum_{n\ge1}a_nx^{n+1}\right)=0$

Consolidate into one series starting $n=1$ :

$\displaystyle2a_2+(6a_3-a_0)x+\sum_{n\ge1}\bigg[(n+3)(n+2)a_{n+3}+(n-1)a_n\bigg]x^{n+1}=0$

Suppose we're given initial conditions $y(0)=a_0$ and $y'(0)=a_1$ (which follow from setting $x=0$ in the power series representations for $y$ and $y'$ , respectively). From the above equation it follows that

$\begin{cases}2a_2=0\\6a_3-a_0=0\\(n+3)(n+2)a_{n+3}+(n-1)a_n=0&\text{for }n\ge2\end{cases}$

Let's first consider what happens when $n=3k-2$ , i.e. $n\in\{1,4,7,10,\ldots\}$ . The recurrence relation tells us that

$a_4=-\dfrac{1-1}{(1+3)(1+2)}a_1=0\implies a_7=0\implies a_{10}=0$

and so on, so that $a_{3k-2}=0$ except for when $k=1$ .

Now let's consider $n=3k-1$ , or $n\in\{2,5,8,11,\ldots\}$ . We know that $a_2=0$ , and from the recurrence it follows that $a_{3k-1}=0$ for all $k$ .

Finally, take $n=3k$ , or $n\in\{0,3,6,9,\ldots\}$ . We have a solution for $a_3$ in terms of $a_0$ , so the next few terms ( $k=2,3,4$ ) according to the recurrence would be

$a_6=-\dfrac2{6\cdot5}a_3=-\dfrac2{6\cdot5\cdot3\cdot2}a_0=-\dfrac{a_0}{6\cdot3\cdot5}$
$a_9=-\dfrac5{9\cdot8}a_6=\dfrac{a_0}{9\cdot6\cdot3\cdot8}$
$a_{12}=-\dfrac8{12\cdot11}a_9=-\dfrac{a_0}{12\cdot9\cdot6\cdot3\cdot11}$

and so on. The reordering of the product in the denominator is intentionally done to make the pattern clearer. We can surmise the general pattern for $n=3k$ as

$a_{3k}=\dfrac{(-1)^{k+1}a_0}{(3k\cdot(3k-3)\cdot(3k-2)\cdot\cdots\cdot6\cdot3\cdot(3k-1)}$
$a_{3k}=\dfrac{(-1)^{k+1}a_0}{3^k(k\cdot(k-1)\cdot\cdots\cdot2\cdot1)\cdot(3k-1)}$
$a_{3k}=\dfrac{(-1)^{k+1}a_0}{3^kk!(3k-1)}$

So the series solution to the ODE is given by

$y=\displaystyle\sum_{n\ge0}a_nx^n$
$y=a_1x+\displaystyle\sum_{k\ge0}\frac{(-1)^{k+1}a_0}{3^kk!(3k-1)}$

Attached is a plot of a numerical solution (blue) to the ODE with initial conditions sampled at $a_0=y(0)=1$ and $a_1=y'(0)=2$ overlaid with the series solution (orange) with $n=3$ and $n=6$ . (Note the rapid convergence.)

7 0

1 year ago