2 times as many as 8 is 20

How to simplify the square root of 3 over the square root of 21

Alika [10]

3 * 7 = 21, so

$\sqrt{21}=\sqrt{3\cdot7}=\sqrt3\sqrt7$

Then

$\dfrac{\sqrt3}{\sqrt{21}}=\dfrac1{\sqrt7}$

and you can rationalize the denominator to write this as

$\dfrac1{\sqrt7}\cdot\dfrac{\sqrt7}{\sqrt7}=\dfrac{\sqrt7}7$

7 0

4 years ago

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4 shirts for $32.00 rate and unit rate

mezya [45]

8 per shirt, 8 per shirt

4 0

4 years ago

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Emilia and Ferdinand took the same freshman chemistry course: Emilia in the fall, Ferdinand in the spring. Emilia made an 83 on

Wewaii [24]

Answer:

$z = \frac{83-76}{8}=0.875$

$z = \frac{79-65}{12}=1.167$

As we can see the z score for Ferdinad is higher than the z score for Emilia so on this case we can conclude that Ferdinand was better compared with his group of reference.

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Emilia case

Let X the random variable that represent the scores of a test, and we know that

Where $\mu=76$ and $\sigma=8$

The z score formula is given by:

$z=\frac{x-\mu}{\sigma}$

Since Emilia made 83 points we can find the z score like this:

$z = \frac{83-76}{8}=0.875$

Ferdinand case

Let X the random variable that represent the scores of a test, and we know that

Where $\mu=65$ and $\sigma=12$

The z score formula is given by:

$z=\frac{x-\mu}{\sigma}$

Since Ferdinand made 79 points we can find the z score like this:

$z = \frac{79-65}{12}=1.167$

As we can see the z score for Ferdinad is higher than the z score for Emilia so on this case we can conclude that Ferdinand was better compared with his group of reference.

5 0

3 years ago

Use the variation of parameters method to solve the DR y" + y' - 2y = 1

postnew [5]

Answer:

$y(t)\ =\ C_1e^{-2t}+C_2e^t-t\dfrac{e^{-2t}}{3}-\dfrac{1}{3}$

Step-by-step explanation:

As given in question, we have to find the solution of differential equation

$y"+y'-2y=1$

by using the variation in parameter method.

From the above equation, the characteristics equation can be given by

$D^2+D-2\ =\ 0$

$=>D=\ \dfrac{-1+\sqrt{1^2+4\times 2\times 1}}{2\times 1}\ or\ \dfrac{-1-\sqrt{1^2+4\times 2\times 1}}{2\times 1}$

$=>\ D=\ -2\ or\ 1$

Since, the roots of characteristics equation are real and distinct, so the complementary function of the differential equation can be by

$y_c(t)\ =\ C_1e^{-2t}+C_2e^t$

Let's assume that

$y_1(t)=e^{-2t}$ $y_2(t)=e^t$

$=>\ y'_1(t)=-2e^{-2t}$ $y'_2(t)=e^t$

and g(t)=1

Now, the Wronskian can be given by

$W=y_1(t).y'_2(t)-y'_1(t).y_2(t)$

$=e^{-2t}.e^t-e^t(-e^{-2t})$

$=e^{-t}+2e^{-t}$

$=3e^{-t}$

Now, the particular solution can be given by

$y_p(t)\ =\ -y_1(t)\int{\dfrac{y_2(t).g(t)}{W}dt}+y_2(t)\int{\dfrac{y_1(t).g(t)}{W}dt}$

$=\ -e^{-2t}\int{\dfrac{e^t.1}{3.e^{-t}}dt}+e^{t}\int{\dfrac{e^{-2t}.1}{3.e^{-t}}dt}$

$=\ -e^{-2t}\int{\dfrac{1}{3}dt}+\dfrac{e^t}{3}\int{e^{-t}dt}$

$=\dfrac{-e^{-2t}}{3}.t-\dfrac{1}{3}$

$=-t\dfrac{e^{-2t}}{3}-\dfrac{1}{3}$

Now, the complete solution of the given differential equation can be given by

$y(t)\ =\ y_c(t)+y_p(t)$

$=C_1e^{-2t}+C_2e^t-t\dfrac{e^{-2t}}{3}-\dfrac{1}{3}$

5 0

3 years ago

The yard behind the Cindy’s house is rectangular in shape and has a perimeter of 72 feet. If the length l of the yard is 18 feet

PtichkaEL [24]

Let
L----------> the length of the yard
W--------> the width l of the yard

we know that
the perimeter is equal to
 $P=2*[W+L]$
$72=2*[W+L] \\ 36=W+L$ --------> equation 1
$L=W+18$ ------> equation 2
substitute equation 2 in equation 1
 $36=W+18+W \\ 36=2W+18 \\ W=(36-18)/2$
$W=9 ft$
$L=W+18 \\ L=9+18 \\ L=27 ft$

we know that
Area of the rectangular yard is equal to
 $A=L*W \\ A=27*9 \\ A=243 ft^{2}$

4 0

4 years ago

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