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2 years ago

13

Horses varied directly as goats and inversely as pigs squared. When the barnyard contained 5 horses there were 4 pigs and only 2

goats. How many goats went with 6 pigs and 10 horses?

1 answer:

Allushta [10]2 years ago

5 0

Answer:

I believe the answer is four there is twice as many horses so you would double the amount of original goats.

You might be interested in

A major traffic problem in the Greater Cincinnati area involves traffic attempting to cross the Ohio River from Cincinnati to Ke

myrzilka [38]

Answer:

a) 0.36

b) 0.3

c) Yes

Step-by-step explanation:

Given:

Probability of no traffic delay in one period, given no traffic delay in the preceding period = P(No_Delay) = 0.9

Probability of finding a traffic delay in one period, given a delay in the preceding period = P(Delay) = 0.6

Period considered = 30 minutes

a)

Let A be the probability that for the next 60 minutes (two time periods) the system will be in the delay state:

As the Probability of finding a traffic delay in one period, given a delay in the preceding period is 0.6 and one period is considered as 30 minutes.

So probability that for the next two time periods i.e. 30*2 = 60 minutes, the system in Delay is

P(A) = P(Delay) * P(Delay) = 0.6 * 0.6 = 0.36

b)

Let B be the probability that in the long run the traffic will not be in the delay state.

This statement means that the traffic will not be in Delay state but be in No_Delay state in long run.

Let C be the probability of one period in Delay state given that preceding period in No-delay state :

P(C) = 1 - P(No_Delay)

= 1 - 0.9

P(C) = 0.1

Now using P(C) and P(Delay) we can compute P(B) as:

P(B) = 1 - (P(Delay) + P(C))

= 1 - ( 0.6 + 0.10 )

= 1 - 0.7

P(B) = 0.3

c)

Yes this assumption should be questioned for this traffic problem because it implies that traffic will be in Delay state for the 30 minutes and just after 30 minutes, it will be in No_Delay state. However, traffic does not work like this in general and it makes this scenario unrealistic. Markov process model can be improved if probabilities are modeled as a function of time instead of being presented as constant (for 30 mins).

8 0

2 years ago

A share of stock in the Lofty Cheese Company is quoted at 25 1/4. Suppose you hold 30 shares of that stock, which you bought at

baherus [9]

Since each share was purchased when it was still quoted at 20 1/4, then when the stock value increases, it gains (25 1/4 - 20 1/4) = $5.00 for each stock. Since you're to sell 30 shares of your stock, that means you'll be making a profit of (30 x 5 ) = $150.00. Thus, the answer is C<span>.</span>

8 0

2 years ago

The perimeter of a pool table is about 7.8m. four times the length equals nine times the width. What are the dimensions of the t

Nana76 [90]

The pool's dimension are l = 2.7 m and w = 1.2 m

Step-by-step explanation:

Step 1 :

Given

The pool's perimeter = 7.8 m

Four times the pool's length = nine times the pool's width.

=> 4l = 9w => l = ( $\frac{9}{4}$ ) w

Where l is the pool's length and w is the pool's width

Step 2:

The rectangle's formula for perimeter is 2 (l+w)

Substituting l = ( $\frac{9}{4}$ ) w ,

2[ ( $\frac{9}{4}$ ) w + w] = 7.8 (given)

( $\frac{13}{2}$ ) w = 7.8 = > w = 1.2 m

Step 3 :

l = ( $\frac{9}{4}$ ) w = ( $\frac{9}{4}$ ) 1.2 = 2.7

l = 2.7 m

Step 4 :

Answer :

The pool's dimension are l = 2.7 m and w = 1.2 m

6 0

2 years ago

An urn contains two blue balls (denoted B1 and B2) and three white balls (denoted W1, W2, and W3). One ball is drawn, its color

alekssr [168]

Answer:

(a) Shown below.

(b) The probability that the first ball drawn is blue is 0.40.

(c) The probability that only white balls are drawn is 0.36.

Step-by-step explanation:

The balls in the urn are as follows:

Blue balls: B₁ and B₂

White balls: W₁, W₂ and W₃

It is provided that two balls are drawn from the urn, with replacement, and their color is recorded.

(a)

The possible outcomes of selecting two balls are as follows:

B₁B₁ B₂B₁ W₁B₁ W₂B₁ W₃B₁

B₁B₂ B₂B₂ W₁B₂ W₂B₂ W₃B₂

B₁W₁ B₂W₁ W₁W₁ W₂W₁ W₃W₁

B₁W₂ B₂W₂ W₁W₂ W₂W₂ W₃W₂

B₁W₃ B₂W₃ W₁W₃ W₂W₃ W₃W₃

There are a total of N = 25 possible outcomes.

(b)

The sample space for selecting a blue ball first is:

S = {B₁B₁, B₁B₂, B₁W₁, B₁W₂, B₁W₃, B₂B₁, B₂B₂, B₂W₁, B₂W₂, B₂W₃}

n (S) = 10

Compute the probability that the first ball drawn is blue as follows:

$P(\text{First ball is Blue})=\frac{n(S)}{N}=\frac{10}{25}=0.40$

Thus, the probability that the first ball drawn is blue is 0.40.

(c)

The sample space for selecting only white balls is:

X = {W₁W₁, W₂W₁, W₃W₁, W₁W₂, W₂W₂, W₃W₂, W₁W₃, W₂W₃, W₃W₃}

n (X) = 9

Compute the probability that only white balls are drawn as follows:

$P(\text{Only White balls})=\frac{n(X)}{N}=\frac{9}{25}=0.36$

Thus, the probability that only white balls are drawn is 0.36.

4 0

2 years ago

The number of cups of sugar (y) in a cookie recipe is proportional to the number of cups of flour (x). The recipe calls for 1.25

8090 [49]

Answer:

Step-by-step explanation:

Y/X = 1.25/1 so place the amount of flour required in the denominator and multiply the denominator amount by 1.25 to find the amount of sugar

This problem appears to be missing equation choices.

6 0

2 years ago

Read 2 more answers