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3 years ago

Find the values of the following:

Mathematics

1 answer:

OLEGan [10]3 years ago

8 0

(1/2)^-2+(1/3)^-2+(1/4)^2

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Z^2+1+y when y=4, and z=3.

OlgaM077 [116]

3^2+1+4=14 the answers 14

7 0

3 years ago

What is the probability that a value falls in the interval μ ± 2σ, that computes the probability P(-2 < Z < 2). Round to f

bija089 [108]

Answer:

0.9544

Step-by-step explanation:

P(-2 < Z < 2) means that Z has mean 0 and standard deviation 2.

P(−2 < Z < 2) = F(2) - F(-2)

Using the Z - table,

F(2) = 0.9772

and F(-2) = 0.0228

Thus,

P(−2 < Z < 2) = 0.9772 - 0.0228 = 0.9544

This means that data within two standard deviation is 95%.

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3 years ago

What fraction does 4 triangles represent when 1 hexagon is worth 1 whole

strojnjashka [21]

Answer: 2/3

Step-by-step explanation: a hexagon is made up of 6 triangles so it would be 4/6 which can be simplified is 2/3

4 0

3 years ago

The projected rate of increase in enrollment at a new branch of the UT-system is estimated by E ′ (t) = 12000(t + 9)−3/2 where E

nexus9112 [7]

Answer:

The projected enrollment is $\lim_{t \to \infty} E(t)=10,000$

Step-by-step explanation:

Consider the provided projected rate.

$E'(t) = 12000(t + 9)^{\frac{-3}{2}}$

Integrate the above function.

$E(t) =\int 12000(t + 9)^{\frac{-3}{2}}dt$

$E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+c$

The initial enrollment is 2000, that means at t=0 the value of E(t)=2000.

$2000=-\frac{24000}{\left(0+9\right)^{\frac{1}{2}}}+c$

$2000=-\frac{24000}{3}+c$

$2000=-8000+c$

$c=10,000$

Therefore, $E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000$

Now we need to find $\lim_{t \to \infty} E(t)$

$\lim_{t \to \infty} E(t)=-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000$

$\lim_{t \to \infty} E(t)=10,000$

Hence, the projected enrollment is $\lim_{t \to \infty} E(t)=10,000$

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3 years ago

Please help with 16 and 17 those are hard

notka56 [123]

The answer to number 17 is c

6 0

3 years ago