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DaniilM [7]
3 years ago
7

The life spans of elephants in a Columbus zoo are normally distributed. The average elephant lives 60 years: the standard deviat

ion is 2 years. Find a range of elephant's life spans which lies within 1 standard deviation, about 68% Between 58 and 62 years Between 56 and 64 years Between 54 and 66 years Between 50 and 60 years
Mathematics
1 answer:
Marta_Voda [28]3 years ago
5 0

did u ever get the answer l need it


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MrRissso [65]

Editing this Answer...

8 0
2 years ago
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Let h(x)=2x−5 and g(x)=−3x+1.Find h(x)+g(x).? help please
Aleksandr-060686 [28]
This may look a little confusing but all you have to do is plug the equation given for h(x) which is 2x-5 in to the h(x) area in the equation where it says h(x) + g(x).

So far that’s (2x-5) + g(x)

Then,

Do the same with equation given for g(x) which is now,

(2x-5) + (3x+1)

Then solve,

2x - 5 + 3x + 1
2x + 3x- 5 + 1
5x - 4

Therefore the answer is 5x - 4.
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3 years ago
Cual ecuacion representa un ejemplo de la
Katarina [22]

Answer:

B) 6(x+2y) + 4 = 6x + 12 y + 4

6 0
3 years ago
Y = 2x + 3<br> y = -3x + 8
snow_lady [41]

Answer:

i did not get that

Step-by-step explanation:

8 0
2 years ago
Evaluate the integral using integration by parts with the indicated choices of u and dv. (Use C for the constant of integration.
trasher [3.6K]

Answer:

\frac{xe^{7x}}{7} + \frac{e^{7x}}{49}

Step-by-step explanation:

Given the integral equation

\int\limits{xe^{7x}} \, dx \\

According to integration by part;

\int\limits {u} \, dv = uv +  \int\limits {v} \, du

u = x, dv = e^7x

du/dx = 1

du = dx

v = \int\limits {e^{7x}} \, dx \\v = e^7x/7

Substitute the given values into the formula;

\int\limits {xe^{7x}} \, dx = x(e^{7x}/7) + \int\limits ({e^{7x}/7}) \, dx\\\int\limits {xe^{7x}} \, dx = \frac{xe^{7x}}{7} + \frac{e^{7x}}{7*7} \\\int\limits {xe^{7x}} \, dx = \frac{xe^{7x}}{7} + \frac{e^{7x}}{49}

3 0
3 years ago
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