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Elanso [62]
3 years ago
12

N(p)=4 means that there a four elements in a set P. Given:

Mathematics
1 answer:
viktelen [127]3 years ago
6 0

Answer:

n(20*u*40=50

Step-by-step explanation:

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Is it proportions or non proportional​
dexar [7]

Answer:

proportional

Step-by-step explanation:

5 0
3 years ago
If sin(y°) = cos(x°), which of the following statements is true?
Firdavs [7]

Answer:

<u>y = w and ΔABC ~ ΔCDE</u>

Step-by-step explanation:

Given sin(y°) = cos(x°)

So, ∠y + ∠x = 90°  ⇒(1)

And as shown at the graph:

ΔABC is aright triangle at B

So, ∠y + ∠z = 90° ⇒(2)

From (1) and (2)

<u>∴ ∠x = ∠z </u>

ΔCDE is aright triangle at D

So, ∠x + ∠w = 90° ⇒(3)

From (1) and (3)

<u>∴ ∠y = ∠w</u>

So, for the triangles ΔABC and ΔCDE

  • ∠A = ∠C  ⇒ proved by ∠y = ∠w
  • ∠B = ∠D  ⇒ Given ∠B and ∠D are right angles.
  • ∠C = ∠E  ⇒ proved by ∠x = ∠z

So, from the previous  ΔABC ~ ΔCDE by AAA postulate.

So, the answer is <u>y = w and ΔABC ~ ΔCDE</u>

4 0
3 years ago
Read 2 more answers
What is the solution to the equation below?<br> log7+log(x-4)= 1
Mila [183]

Answer: x=38/7

Step-by-step explanation:

8 0
3 years ago
Each of the 6 cats in a pet store was weighed. Here are their weights
zhannawk [14.2K]

Answer:

Mean= 10.83

Median= 9.5

Step-by-step explanation:

Given number,

10, 15, 8, 11, 9, 12

N= 6

As we know,

Mean= Ex/N

=10+15+8+11+9+12/6

=65/6

=10.83

Therefore, Mean= 10.83

Now,

Median(MD)= (N+1/2)th item

= 6+1/2

=7/2

=3.5th item

So, Median(MD)= 3th+4th/2 item

= 8+11/2

= 19/2

= 9.5

Therefore, Median= 9.5

7 0
3 years ago
The population of a community is known to increase at a rate proportional to the number of people present at time t. The initial
qaws [65]

Answer:

Step-by-step explanation:

Let P be the population of the community

So the population of a community increase at a rate proportional to the number of people present at a time

That is

\frac{dp}{dt} \propto p\\\\\frac{dp}{dt} =kp\\\\ [k \texttt {is constant}]\\\\\frac{dp}{dt} -kp =0

Solve this equation we get

p(t)=p_0e^{kt}---(1)

where p is the present population

p₀ is the initial population

If the  initial population as doubled in 5 years

that is time t = 5 years

We get

2p_o=p_oe^{5k}\\\\e^{5k}=2

Apply In on both side to get

Ine^{5k}=In2\\\\5k=In2\\\\k=\frac{In2}{5} \\\\\therefore k=\frac{In2}{5}

Substitute k=\frac{In2}{5}  in p(t)=p_oe^{kt} to get

\large \boxed {p(t)=p_oe^{\frac{In2}{5}t }}

Given that population of a community is 9000 at 3 years

substitute t = 3 in {p(t)=p_oe^{\frac{In2}{5}t }}

p(3)=p_oe^{3 (\frac{In2}{5}) }\\\\9000=p_oe^{3 (\frac{In2}{5}) }\\\\p_o=\frac{9000}{e^{3(\frac{In2}{5} )}} \\\\=5937.8

<h3>Therefore, the initial population is 5937.8</h3>
7 0
3 years ago
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