......................................pic below.............................. subject is math

hat is the speed of a wave if the wavelength is 18 m and the frequency is 35 Hz? (Quick Hurry, need answer)

Andrew [12]

The speed is 630mp/s I think. Because you have to multiply Wavelength by Frequency to get true speed

3 0

3 years ago

How many 5-member chess teams can be chosen from 15 interested players? Consider only the members selected, not their board posi

ryzh [129]

Answer:

3003 number of 5-member chess teams can be chosen from 15 interested players.

Step-by-step explanation:

Given:

Number of the interested players = 15

To Find:

Number of 5-member chess teams that can be chosen = ?

Solution:

Combinations are a way to calculate the total outcomes of an event where order of the outcomes does not matter. To calculate combinations, we will use the formula $nCr = \frac{n!}{r!(n - r)!}$

where

n represents the total number of items,

r represents the number of items being chosen at a time.

Now we have n = 15 and r = 5

Substituting the values,

$15C_5 = \frac{15!}{5!(15- 5)!}$

$15C_5 = \frac{15!}{5!(10)!}$

$15C_5 = \frac{15\times \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 }{5 \times 4 \times 3 \times 2 \times 1(10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)}$

$15C_5 = \frac{15\times \times 14 \times 13 \times 12 \times 11}{(5 \times 4 \times 3 \times 2 \times 1)}$

$15C_5 = \frac{360360}{120}$

$15C_5 = 3003$

7 0

3 years ago

Express 5.1146• 10^3 in standard notation

charle [14.2K]

Answer:

5,114.6

Step-by-step explanation:

5.1146 x 10^3 means move decimal point 3 to the right to get the answer

4 0

2 years ago

A given field mouse population satisfies the differential equation dp dt = 0.5p − 410 where p is the number of mice and t is the

ohaa [14]

Answer:

a) $t = 2 *ln(\frac{82}{5}) =5.595$

b) $t = 2 *ln(-\frac{820}{p_0 -820})$

c) $p_0 = 820-\frac{820}{e^6}$

Step-by-step explanation:

For this case we have the following differential equation:

$\frac{dp}{dt}=\frac{1}{2} (p-820)$

And if we rewrite the expression we got:

$\frac{dp}{p-820}= \frac{1}{2} dt$

If we integrate both sides we have:

$ln|P-820|= \frac{1}{2}t +c$

Using exponential on both sides we got:

$P= 820 + P_o e^{1/2t}$

Part a

For this case we know that p(0) = 770 so we have this:

$770 = 820 + P_o e^0$

$P_o = -50$

So then our model would be given by:

$P(t) = -50e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=-50 e^{1/2 t} +820$

$\frac{820}{50} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(\frac{82}{5}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(\frac{82}{5}) =5.595$

Part b

For this case we know that p(0) = p0 so we have this:

$p_0 = 820 + P_o e^0$

$P_o = p_0 -820$

So then our model would be given by:

$P(t) = (p_o -820)e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=(p_o -820)e^{1/2 t} +820$

$-\frac{820}{p_0 -820} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(-\frac{820}{p_0 -820}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(-\frac{820}{p_0 -820})$

Part c

For this case we want to find the initial population if we know that the population become extinct in 1 year = 12 months. Using the equation founded on part b we got:

$12 = 2 *ln(\frac{820}{820-p_0})$