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Morgarella [4.7K]
2 years ago
11

. In an extra-curricular club with 15 members,7 people played rugby, 6 people played soccer and 4 people neither play rugby nor

soccer. How many people played both rugby and soccer?
Mathematics
1 answer:
inna [77]2 years ago
7 0

Answer:

2

Step-by-step explanation:

In the above question, we are given the following information:

Total member in the club = 15

Rugby = n(R) = 7

Soccer = n(S) = 6

Neither Rugby nor Soccer = 4

Rugby and soccer = n( R ∩ S) = (Unknown)

Total number of club members = n(R) + n(S) - n( R ∩ S) + Neither Rugby nor soccer

15 = 7 + 6 - n( R ∩ S) + 4

15 = 17 - n( R ∩ S)

15 - 17 = - n( R ∩ S)

-2 = - n( R ∩ S)

n( R ∩ S) = 2

Therefore, the number of people that played both rugby and soccer is 2

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Kyle challenges you to a game. Each player rolls a standard number cube twice. If the sum of the results is divisible by 3, you
sweet [91]

Answer:

The probability of you winning is 1/3

No, it is not a fair game

Step-by-step explanation:

The first thing we need to have here is the sample space. This refers to the set of all possible results that can occur from the rolling.

Please check attachment for this

Kindly note that the total number of possible outcomes is 36.

And in the attachment, sums which are divisible by 3 are circled.

The number of circles we can count is 12

Thus, the probability of you winning would be number of circles/total number of outcomes = 12/36 = 1/3

Is it a fair game?

No, it is not

It can only be a fair game if the probability of winning equals probability of losing ( which is 18/36 = 1/2 or 0.5)

8 0
2 years ago
A small ferryboat is 4.00 m wide and 6.00 m long. When a loaded truck pulls onto it, the boat sinks an additional 4.00 cm into t
Svet_ta [14]

Answer:

F_w=9408\ N

Weight of the truck=9408 N

Step-by-step explanation:

Boat is experiencing the buoyant force as it is in the water and is sinking

According to the force balance in y direction. As both is floating, two forces balance each other:

F_b-F_w=0

where:

F_b is the buoyant force

F_w is the weight=mg

F_b=F_w        Eq (1)

Buoyant force is equal to the mass of water displaced * gravitational acceleration.

F_b=m_{water\ displaced}*g\\F_b=\rho Vg\\

Taking density of water to be 1000 Kg/m^3

F_b=1000*(6*4*4*10^{-2})*9.8\\F_b=9408 N

From Eq(1):

F_w=9408\ N

Weight of the truck=9408 N

6 0
2 years ago
If y varies directly as x, and y is 400 when x is r and y is r when x is 4, what is the numeric constant of variation in this
otez555 [7]

Answer:

10

Step-by-step explanation:

If there is a direct relation between two variables x and y then it can be represented as

y = kx ,

where y is dependent variable

x is  independent variable

k is constant of variation

_____________________________

First condition

y = 400

x = r

using y = kx then relationship will be

400 = kr

finding k here

k = 400/r

Second condition

y = r

x = 4

using y = kx then relationship will be

r = 4k

finding k

k = r/4

since  in both condition equation is same

thus, value of k will also be same

thus,

400/r = r/4

=> 400*4 = r*r

=> 1600 = r^2

\sqrt{r^{2} }  = \sqrt{1600} \\r = 40

Thus, 40 is the value of r

k = r/4 = 40/4 = 10

Thus, constant of variation is 10 which is correct choice.

To cross validate

k = 400/r = 400/40 = 10

7 0
1 year ago
Read 2 more answers
Preston is making cookies. He has fraction 4 over 5 cup of flour in a bag. He then used two fraction 1 over 10 cup scoops of flo
Art [367]

I'm pretty sure the answer will be 6/10





Hope this helps, best of luck, terribly sorry if I'm wrong if I am my  apologies




~Animaljamissofab ♥

5 0
2 years ago
Read 2 more answers
Juan invest $3700 in a simple interest account at a rate of 4% for 15 years
OleMash [197]

<em><u>Question:</u></em>

Juan Invest $3700 In A Simple Interest Account At A Rate Of 4% For 15 Years. How Much Money Will Be In The Account After 15 Years?

<em><u>Answer:</u></em>

There will be $ 5920 in account after 15 years

<em><u>Solution:</u></em>

<em><u>The simple interest is given by formula:</u></em>

S.I = \frac{p \times n \times r }{100}

Where,

p is the principal

n is number of years

r is rate of interest

From given,

p = 3700

r = 4 %

t = 15 years

Therefore,

S.I = \frac{3700 \times 4 \times 15 }{100}\\\\S.I = 37 \times 4 \times 15\\\\S.I = 2220

<em><u>How Much Money Will Be In The Account After 15 Years?</u></em>

Total money = principal + simple interest

Total money = 3700 + 2220

Total money = 5920

Thus there will be $ 5920 in account after 15 years

8 0
2 years ago
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