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Basile [38]
3 years ago
11

Use the Pythagorean theorem to find the length of the diagonal of a square that has

Mathematics
1 answer:
worty [1.4K]3 years ago
3 0

Answer:

How far (nearest foot) will it be from home plate to second base, assuming he builds it to that specification? ... A square lot has an area of 1,200 square meters. .... Therefore, we can use the Pythagorean Theorem to solve for the diagonal:.

Step-by-step explanation:

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In the figure below what is the name of the angle formed by two rays QR and QP? What is the common. endpoint,for this angle?
castortr0y [4]
The common endpoint is Q. The angle is: angle RQP, angle PQR, or just angle Q
6 0
3 years ago
Imagine an experiment having three conditions and 20 subjects within each condition. The mean and variances of each condition ar
xxTIMURxx [149]

Answer:

1. Mean square B= 5.32

2. Mean square E= 16.067

3. F= 0.33

4. p-value: 0.28

Step-by-step explanation:

Hello!

You have the information of 3 groups of people.

Group 1

n₁= 20

X[bar]₁= 3.2

S₁²= 14.3

Group 2

n₂= 20

X[bar]₂= 4.2

S₂²= 17.2

Group 3

n₃= 20

X[bar]₃= 7.6

S₃²= 16.7

1. To manually calculate the mean square between the groups you have to calculate the sum of square between conditions and divide it by the degrees of freedom.

Df B= k-1 = 3-1= 2

Sum Square B is:

∑ni(Ÿi - Ÿ..)²

Ÿi= sample mean of sample i ∀ i= 1,2,3

Ÿ..= general mean is the mean that results of all the groups together.

General mean:

Ÿ..= (Ÿ₁ + Ÿ₂ + Ÿ₃)/ 3 = (3.2+4.2+7.6)/3 = 5

Sum Square B (Ÿ₁ - Ÿ..)² + (Ÿ₂ - Ÿ..)² + (Ÿ₃ - Ÿ..)²= (3.2 - 5)² + (4.2 - 5)² + (7.6 - 5)²= 10.64

Mean square B= Sum Square B/Df B= 10.64/2= 5.32

2. The mean square error (MSE) is the estimation of the variance error (σ_{e}^2 → S_{e} ^{2}), you have to use the following formula:

Se²=<u> (n₁-1)S₁² + -(n₂-1)S₂² + (n₃-1)S₃²</u>

                        n₁+n₂+n₃-k

Se²=<u> 19*14.3 + 19*17.2 + 19*16.7 </u>= <u>  915.8   </u>  = 16.067

                 20+20+20-3                  57

DfE= N-k = 60-3= 57

3. To calculate the value of the statistic you have to divide the MSB by MSE

F= \frac{Mean square B}{Mean square E} = \frac{5.32}{16.067} = 0.33

4. P(F_{2; 57} ≤ F) = P(F_{2; 57} ≤ 0.33) = 0.28

I hope you have a SUPER day!

3 0
3 years ago
On a number line, a number, b, is located the same distance from 0 as another number, a, but in the opposite direction. The numb
vfiekz [6]

Answer:

b

Step-by-step explanation:

5 0
3 years ago
The variable z is directly proportional to x, and inversely proportional to y. When x is 6 and y is 6, z has the value 12.
dalvyx [7]

z is directly proportional to x: z = kx.

If z is directly proportional to x and inversely proportional to y, then z = Kx/y.

When x= 6 and y = 6, z is 12. Find the constant of proportionality, K:

12 = K(6/6), or K = 12

Then, in general, z = 12x/y.

What is z when x=13 and y=10? z = 12(13)/(10) = 15.6 (answer)

3 0
3 years ago
Please I need help with differential equation. Thank you
Inga [223]

1. I suppose the ODE is supposed to be

\mathrm dt\dfrac{y+y^{1/2}}{1-t}=\mathrm dy(t+1)

Solving for \dfrac{\mathrm dy}{\mathrm dt} gives

\dfrac{\mathrm dy}{\mathrm dt}=\dfrac{y+y^{1/2}}{1-t^2}

which is undefined when t=\pm1. The interval of validity depends on what your initial value is. In this case, it's t=-\dfrac12, so the largest interval on which a solution can exist is -1\le t\le1.

2. Separating the variables gives

\dfrac{\mathrm dy}{y+y^{1/2}}=\dfrac{\mathrm dt}{1-t^2}

Integrate both sides. On the left, we have

\displaystyle\int\frac{\mathrm dy}{y^{1/2}(y^{1/2}+1)}=2\int\frac{\mathrm dz}{z+1}

where we substituted z=y^{1/2} - or z^2=y - and 2z\,\mathrm dz=\mathrm dy - or \mathrm dz=\dfrac{\mathrm dy}{2y^{1/2}}.

\displaystyle\int\frac{\mathrm dy}{y^{1/2}(y^{1/2}+1)}=2\ln|z+1|=2\ln(y^{1/2}+1)

On the right, we have

\dfrac1{1-t^2}=\dfrac12\left(\dfrac1{1-t}+\dfrac1{1+t}\right)

\displaystyle\int\frac{\mathrm dt}{1-t^2}=\dfrac12(\ln|1-t|+\ln|1+t|)+C=\ln(1-t^2)^{1/2}+C

So

2\ln(y^{1/2}+1)=\ln(1-t^2)^{1/2}+C

\ln(y^{1/2}+1)=\dfrac12\ln(1-t^2)^{1/2}+C

y^{1/2}+1=e^{\ln(1-t^2)^{1/4}+C}

y^{1/2}=C(1-t^2)^{1/4}-1

I'll leave the solution in this form for now to make solving for C easier. Given that y\left(-\dfrac12\right)=1, we get

1^{1/2}=C\left(1-\left(-\dfrac12\right)^2\right))^{1/4}-1

2=C\left(\dfrac54\right)^{1/4}

C=2\left(\dfrac45\right)^{1/4}

and so our solution is

\boxed{y(t)=\left(2\left(\dfrac45-\dfrac45t^2\right)^{1/4}-1\right)^2}

3 0
3 years ago
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