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3 years ago

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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

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