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Len [333]
2 years ago
12

The mean high temperature on a particular day in January is 31 degrees F and the standard deviation is 8.7 degrees. One year, th

e temperature was 52 degrees F on that day. What is the Z-score for that day's temperature? Round to the appropriate number of decimal places for Z-scores. Is this temperature significantly high? (greater than 2)​
Mathematics
1 answer:
marin [14]2 years ago
4 0

Answer:

jsjs gnckdoesosensjkcjbdjd

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Please help, test due in 1 hour! I will give brainliest! :)
Aleksandr-060686 [28]

Answer:

Rectangle + half circle ?

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
In Exercises 3–10, use the Chain Rule to calculate the partial derivatives. Express the answer in terms of the independent varia
sashaice [31]

I'll use subscript notation for brevity, i.e.

\dfrac{\partial f}{\partial x}=f_x

3.

f(x,y,z)=xy+z^2\implies\begin{cases}f_x=y\\f_y=x\\f_z=2z\end{cases}

x(r,s)=s^2\implies\begin{cases}x_r=0\\x_s=2s\end{cases}

y(r,s)=2rs\implies\begin{cases}y_r=2s\\y_s=2r\end{cases}

z(r,s)=r^2\implies\begin{cases}z_r=2r\\z_s=0\end{cases}

By the chain rule,

f_r=f_xx_r+f_yy_r+f_zz_r=2xs+4zr=2s^3+4r^3

f_s=f_xx_s+f_yy_s+f_zz_s=2ys+2xr=6rs^2

4.

x(r,s,t)=r+s-2t\implies\begin{cases}x_r=1\\x_s=1\\x_t=-2\end{cases}

y(r,s,t)=3rt\implies\begin{cases}y_r=3t\\y_s=0\\y_t=3r\end{cases}

z(r,s,t)=s^2\implies\begin{cases}z_r=0\\z_s=2s\\z_t=0\end{cases}

By the chain rule,

f_r=f_xx_r+f_yy_r+f_zz_r=y+3xt=3rt+3r+3s-6t^2

f_s=f_xx_s+f_yy_s+f_zz_s=y+4zs=3rt+4s^3

f_t=f_xx_t+f_yy_t+f_zz_t=-2y+3xr=3r+3s-12rt

8 0
2 years ago
The mass of an object is x^15 grams. Its volume is x^9 cm^3. What is the object’s density?
VashaNatasha [74]

Answer:

Step-by-step explanation:

density is mass divided by volume

15 divided by 9 is

1.6666666666

so its 1.666 repeating g/cm^3

8 0
1 year ago
NEED ANSWER ASAP
Ivenika [448]

Answer:

D

Step-by-step explanation:11y+17=11y-10\\17=/=-10\\

17 does not equal -10 therefore it is no solution

8 0
2 years ago
What is the answer to C??
Licemer1 [7]

Answer:

x = 1

Step-by-step explanation:

3^2x - 9 (3^-2x) = 8

<=> 3^2x - 9/(3^2x) = 8

Given that 3^2x = y [1] (y>0), we have a new equation as follows:

y - 9×\frac{1}{y} = 8

<=> y - 9/y = 8

<=> y(y - 9/y) = y×8

<=> y^2 - 9 = 8×y

<=> y^2 - 8y - 9 = 0

<=> (y^2 + y) - (9×y + 9) = 0

<=> y(y + 1) - 9(y + 1) = 0

<=> (y - 9)(y + 1) = 0

=> y = 9, or y = - 1

However, as 3^2x > 0, y>0.

Therefore, y = 9.

Replace y with 9 in [1], we have:

3^2x = 9 = 3^2

=> 2x = 2

=> x = 1

4 0
2 years ago
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