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3 years ago

Could somebody help.

Mathematics

1 answer:

Marianna [84]3 years ago

8 0

The answer is 4/9.

To find the answer you would need to find the values of x and y. Using the given information you know x/5 is equal to 2/3. Then you would cross multiply and divide in order to fine the value of x, which is 10/3. Doing the same thing to find y, which would be 15/2. Since it’s day the ratio of x and y. You would divide the value of x by y to get your answer of 4/9.

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Mr.holms used 4/5 of a carton of orange juice.he used equal amounts of leftover juice for two servings what fraction of the whol

Sunny_sXe [5.5K]

He used 8/5 of juice there you go

8 0

3 years ago

The ice cream man just ended his shift for the day. Let 1/2x^2 6/11x + 8 represent the amount of chocolate ice cream bars he sol

saveliy_v [14]

Answer:

A and D

Step-by-step explanation:

Total ice cream bars sold = sum of chocolate sold , vanilla and strawberry ice-creams sold.

=(1/2)x2 + (6/11)x + 8 + (5/9)x2 + (2/3) +(1/3)x2 + 4x +(4/3) (Given in the question)

=(25/18)x2 + (50/11)x + 10 (Adding terms corresponding to x2,x ,constant respectively)

Difference in chocolate and strawberry bars =[ (1/2)x2 + (6/11)x + 8] - [(1/3)x2 + 4x +(4/3)]

= (1/6)x2 - (38/11)x +(20/3)

So, the correct options are A and D

4 0

3 years ago

What is the length of RS.

Natali [406]

Consider, in ΔRPQ,

RP = R (Radius of larger circle)

PQ = r (radius of smaller circle)

We have to find, RQ, by Pythagoras theorem,

RP² = PQ²+RQ²

R² = r²+RQ²

RQ² = R²-r²

RQ = √(R²-r²

Now, as RQ & QS both are tangents of the smaller circle, their lengths must be equal. so, RS = 2 × RQ

RS = 2√(R²-r²)

5 0

3 years ago

We have two fair three-sided dice, indexed by i = 1, 2. Each die has sides labeled 1, 2, and 3. We roll the two dice independent

Bogdan [553]

Answer:

(a) P(X = 0) = 1/3

(b) P(X = 1) = 2/9

(d) P(X = 3) = 0

(a) P(Y = 0) = 0

(b) P(Y = 1) = 1/3

Step-by-step explanation:

Given:

- Two 3-sided fair die.

- Random Variable X_1 denotes the number you get for rolling 1st die.

- Random Variable X_2 denotes the number you get for rolling 2nd die.

- Random Variable X = X_2 - X_1.

Solution:

- First we will develop a probability distribution of X such that it is defined by the difference of second and first roll of die.

- Possible outcomes of X : { - 2 , -1 , 0 ,1 , 2 }

- The corresponding probabilities for each outcome are:

( X = -2 ): { X_2 = 1 , X_1 = 3 }

P ( X = -2 ): P ( X_2 = 1 ) * P ( X_1 = 3 )

: ( 1 / 3 ) * ( 1 / 3 )

: ( 1 / 9 )

( X = -1 ): { X_2 = 1 , X_1 = 2 } + { X_2 = 2 , X_1 = 3 }

P ( X = -1 ): P ( X_2 = 1 ) * P ( X_1 = 3 ) + P ( X_2 = 2 ) * P ( X_1 = 3)

: ( 1 / 3 ) * ( 1 / 3 ) + ( 1 / 3 ) * ( 1 / 3 )

: ( 2 / 9 )

( X = 0 ): { X_2 = 1 , X_1 = 1 } + { X_2 = 2 , X_1 = 2 } + { X_2 = 3 , X_1 = 3 }

P ( X = -1 ):P ( X_2 = 1 )*P ( X_1 = 1 )+P( X_2 = 2 )*P ( X_1 = 2)+P( X_2 = 3 )*P ( X_1 = 3)

: ( 1 / 3 ) * ( 1 / 3 ) + ( 1 / 3 ) * ( 1 / 3 ) + ( 1 / 3 ) * ( 1 / 3 )

: ( 3 / 9 ) = ( 1 / 3 )

( X = 1 ): { X_2 = 2 , X_1 = 1 } + { X_2 = 3 , X_1 = 2 }

P ( X = 1 ): P ( X_2 = 2 ) * P ( X_1 = 1 ) + P ( X_2 = 3 ) * P ( X_1 = 2)

: ( 1 / 3 ) * ( 1 / 3 ) + ( 1 / 3 ) * ( 1 / 3 )

: ( 2 / 9 )

( X = 2 ): { X_2 = 1 , X_1 = 3 }

P ( X = 2 ): P ( X_2 = 3 ) * P ( X_1 = 1 )

: ( 1 / 3 ) * ( 1 / 3 )

: ( 1 / 9 )

- The distribution Y = X_2,

P(Y=0) = 0

P(Y=1) = 1/3

P(Y=2) = 1/ 3

- The probability for each number of 3 sided die is same = 1 / 3.

7 0

3 years ago

When a person is breathing normally the amount of air in their lawns varies sinusoidally. When full Karen’s lungs hold 2.8 L of

makkiz [27]

Answer:

$A(t) = 2.2\sin \frac{(t - 2)\pi }{6} + 0.6$

Step-by-step explanation:

Let the function of quantity in the lung of air be A(t)

So $A(t) \alpha \sin (\frac{t - \alpha }{k} )$

so, A(t) = Amax sin t + b

A(t) = 2.8t⇒ max

A(t) = 0.6t ⇒ min

max value of A(t) occur when sin(t) = 1

and min value of A(t) = 0

So b = 0.6

and A(max) = 2.2

$A(t) = 2.2\sin \frac{(t)}{k} + 0.6$

at t = 2 sec volume of a is 0.6

So function reduce to

$A(t) = 2.2\sin \frac{(t - 2)}{k} + 0.6$

and t = 5 max value of volume is represent

so,

$\sin \frac{t - \alpha }{k} = 1$

$\frac{t - 2}{k} = \frac{\pi }{2}$ when t = 5

$\frac{6}{\pi } = k$

so the equation becomes

$A(t) = 2.2\sin \frac{(t - 2)\pi }{6} + 0.6$

7 0

3 years ago

Could somebody help.​

Could somebody help.