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DaniilM [7]
3 years ago
8

thirty percent of those in attendance have never seen a lacrosse match before. how many of the 400 in attendance were watching t

heir first game of lacrosse? how many have previously seen a lacrosse game
Mathematics
1 answer:
Maurinko [17]3 years ago
7 0

Answer:

sorry can u give me a like im tring to get points it my first day

Step-by-step explanation:

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Pita​ Corporation's management has budgeted the following amounts for its next fiscal​ year: Total fixed expenses $490,000 Selli
Ede4ka [16]

Answer:

let me tell you firstly i am juat writing this

Step-by-step explanation:

for absolutely no rwason whatsoevwr lol

ao ywah get some hwlp from a pro

3 0
3 years ago
A standard deck of playing cards has 52 cards total that contains 13 of each suit (hearts, diamonds, clubs and spades). What is
schepotkina [342]

Answer:

C 1/2

Step-by-step explanation:

There are 4 suits, 2 suits are  red (hearts and diamonds) while 2 are black (clubs and spades)

Since 13 cards are in each suit, 26 cards are red ( 2 * 13)  

There are 52 total cards

P (red) = red cards/ total cards

           = 26 / 52

           = 1/2

8 0
3 years ago
The probability of an event is 3/10 . What are the odds of the same event?
kozerog [31]

Answer:

D. \frac{3}{7}

Step-by-step explanation:

We have been given that the probability of an event is 3/10.  

To find the odds of the same event we will use formula:

\text{Odds of an event}=\frac{\text{Probability of the event}}{\text{1-Probability of the event}}

\text{Odds of the event}=\frac{\frac{3}{10}}{1-\frac{3}{10}}

\text{Odds of the event}=\frac{\frac{3}{10}}{\frac{1*10}{10}-\frac{3}{10}}

\text{Odds of the event}=\frac{\frac{3}{10}}{\frac{10-3}{10}}

\text{Odds of the event}=\frac{\frac{3}{10}}{\frac{7}{10}}

Dividing a fraction with another fraction is same as multiplying the 1st fraction by the reciprocal of second fraction.

\text{Odds of the event}=\frac{3}{10}\times \frac{10}{7}

\text{Odds of the event}=\frac{3}{7}

Therefore, the odds of the same event is \frac{3}{7} and option D is the correct choice.


8 0
3 years ago
Need help please!!!!!!
zlopas [31]
The distance from a to c is ten units
8 0
3 years ago
In quadratic drag problem, the deceleration is proportional to the square of velocity
Mars2501 [29]
Part A

Given that a= \frac{dv}{dt} =-kv^2

Then, 

\int dv= -kv^2\int dt \\  \\ \Rightarrow v(t)=-kv^2t+c

For v(0)=v_0, then

v(0)=-kv^2(0)+c=v_0 \\  \\ \Rightarrow c=v_0

Thus, v(t)=-kv(t)^2t+v_0

For v(t)= \frac{1}{2} v_0, we have

\frac{1}{2} v_0=-k\left( \frac{1}{2} v_0\right)^2t+v_0 \\  \\ \Rightarrow \frac{1}{4} kv_0^2t=v_0- \frac{1}{2} v_0= \frac{1}{2} v_0 \\  \\ \Rightarrow kv_0t=2 \\  \\ \Rightarrow t= \frac{2}{kv_0}


Part B

Recall that from part A, 

v(t)= \frac{dx}{dt} =-kv^2t+v_0 \\  \\ \Rightarrow dx=-kv^2tdt+v_0dt \\  \\ \Rightarrow\int dx=-kv^2\int tdt+v_0\int dt+a \\  \\ \Rightarrow x=- \frac{1}{2} kv^2t^2+v_0t+a

Now, at initial position, t = 0 and v=v_0, thus we have

x=a

and when the velocity drops to half its value, v= \frac{1}{2} v_0 and t= \frac{2}{kv_0}

Thus,

x=- \frac{1}{2} k\left( \frac{1}{2} v_0\right)^2\left( \frac{2}{kv_0} \right)^2+v_0\left( \frac{2}{kv_0} \right)+a \\  \\ =- \frac{1}{2k} + \frac{2}{k} +a

Thus, the distance the particle moved from its initial position to when its velocity drops to half its initial value is given by

- \frac{1}{2k} + \frac{2}{k} +a-a \\  \\ = \frac{2}{k} - \frac{1}{2k} = \frac{3}{2k}
7 0
3 years ago
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