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Alecsey [184]
3 years ago
8

How do i solve this problem

Mathematics
1 answer:
Lorico [155]3 years ago
5 0
9.10075 x 10^{7}
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Find the mean of 11, – 7, – 14, 10, and -5.
Rudik [331]

Answer:

-1

Step-by-step explanation:

6 0
3 years ago
A population has the following characteristics. (a) A total of 75% of the population survives the first year. Of that 75%, 25% s
Artemon [7]

Answer:

= \left[\begin{array}{ccc}1344\\84\\28\end{array}\right]  \left \begin{array}{ccc}{0 \  \leq  age   \leq  1 }\\{ 1 \  \leq  age   \leq  2 }\\{2 \  \leq  age  \leq 3}\end{array}\right

i.e after the first year ;

there 1344 members in the first age class

84 members for the second age class; and

28 members for the third age class

Step-by-step explanation:

We can deduce that the age distribution vector x represents the number of population members for each age class; Given that in each class of age there are 112 members present.

The current age distribution vector is as follows:

x = \left[\begin{array}{ccc}1&1&2\\1&1&2\\1&1&2\end{array}\right] \left[\begin{array}{ccc}{0 \  \leq  age   \leq  1 }\\{ 0 \  \leq  age   \leq  2 }\\{0 \  \leq  age   \leq 3}\end{array}\right]

Also , the age transition matrix is as follows:

L = \left[\begin{array}{ccc}3&6&3\\0.75&0&0 \\0&0.25&0\end{array}\right]

After 1 year ; the age distribution vector will be :

x_2 =Lx_1 = \left[\begin{array}{ccc}3&6&3\\0.75&0&0 \\0&0.25&0\end{array}\right]  \left[\begin{array}{ccc}1&1&2\\1&1&2\\1&1&2\end{array}\right]

= \left[\begin{array}{ccc}1344\\84\\28\end{array}\right]  \left \begin{array}{ccc}{0 \  \leq  age   \leq 1 }\\{ 1 \  \leq  age   \leq  2 }\\{2 \  \leq  age   \leq  3}\end{array}\right

6 0
3 years ago
13 If CDEF is a rhombus, find m FED.
Tatiana [17]

Answer:

36

Step-by-step explanation:

8x-20=5x+1

-5       -5

3x=-20+1

+20 +20

3x=21

3  3

x=7

8(7)-20

FED=36

8 0
2 years ago
Consider a series circuit consisting of a resistor of R ohms, an inductor of L henries, and variable voltage source of V(t) volt
ser-zykov [4K]

Answer:

I(t)=\frac{1}{3}(1-e^{-30t})

Step-by-step explanation:

We are given that

\frac{dI}{dt}+\frac{R}{L}I=\frac{V(t)}{L}

R=150 ohm

L=5 H

V(t)=10 V

P=\frac{R}{L}=\frac{150}{5}=30

I.F=e^{\int Pdt}=e^{\int 30 dt}=e^{30 t}

I(t)\times I.F=\int e^{30 t}\times 10 dt+C

I(t)\times e^{30 t}=\frac{10}{30}e^{30 t}+C

I(t)=\frac{1}{3}+Ce^{-30 t}

I(0)=0

Substitute t=0

0=\frac{1}{3}+C

C=-\frac{1}{3}

Substitute the values

I(t)=\frac{1}{3}-\frac{1}{3}e^{-30 t}

I(t)=\frac{1}{3}(1-e^{-30t})

5 0
3 years ago
Danny had 20 minutes to do a three-problem quiz. He spent 9 1/4 minutes on question A and 5 4/5 minutes on question B. How much
Travka [436]

Answer:

4 19/20

Step-by-step explanation:

9 1/4 + 5 4/5 = 15 1/20

20- 15 1/20 = 4  19/20

6 0
2 years ago
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