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2 years ago

5

At a dairy farm the ratio of adult milking cows to young calves was 9:2. If there were 234 adult milking cows at the farm, how m

any young calves were there?

1 answer:

jok3333 [9.3K]2 years ago

7 0

Answer:

52 young calves

Step-by-step explanation:

At a dairy farm the ratio of adult milking cows to young calves was 9:2. If there were 234 adult milking cows at the farm, how many young calves were there?

The ration is given as:

adult milking cows to young calves

= 9 : 2

= 234 : x

= 9/2 = 234/x

x = Number of young calves

Cross Multiply

9 × x = 2 × 234

= x = 2 × 234/9

x = 468/9

x = 52

Hence, the number of young calves on the farm is 52.

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Each morning, Hungry Harry eats some eggs. On any given morning, the number of eggs he eats is equally likely to be 1, 2, 3, 4,

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Answer:

Mean = 35

Variance = 291.7

Step-by-step explanation:

Data provided in the question:

X : 1, 2, 3, 4, 5, 6

All the data are independent

Thus,

The mean for this case will be given as:

Mean, E[X] = $\frac{\textup{Sum of all the observations}}{\textup{Total number of observations}}$

or

E[X] = $\frac{\textup{1+2+3+4+5+6}}{\textup{6}}$

or

E[X] = 3.5

For 10 days, Mean = 3.5 × 10 = 35

And,

variance = E[X²] - ( E[X] )²

Now, for this case of independent value,

E[X²] = $\frac{1^2+2^2+3^2+4^2+5^2+6^2}{\textup{6}}$

or

E[X²] = $\frac{1+4+9+16+25+36}{\textup{6}}$

or

E[X²] = $\frac{91}{\textup{6}}$

or

E[X²] = 15.167

Therefore,

variance = E[X²] - ( E[X] )²

or

variance = 15.167 - 3.5²

or

Variance = 2.917

For 10 days = Variance × Days²

= 2.917 × 10²

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3 years ago

Write an equation in point-slope form of the line that passes through

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This one was though, but I think it might be y= -1/12x - 53/12.

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3 years ago

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Person A can make a cupcake in 5 minutes, Person B can make a cupcake in 9 minutes. In total they made 168 cupcakes. How many cu

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Answer:

Person A made 108 cupcakes and person B made 60 cupcakes.

Step-by-step explanation:

Person A makes 1 cupcake in 5 minutes.

Say in x minutes A makes : $\frac{x}{5}$ cupcakes

Person B makes 1 cupcake in 9 minutes.

Say in x minutes B makes : $\frac{x}{9}$ cupcakes

Total cupcakes prepared by A and B in x minutes = 168

$\frac{x}{5}+\frac{x}{9}=168$

$\frac{14x}{45}=168$

$x=\frac{168\times 45}{14}=540$

So, cupcakes made by A = $\frac{540}{5}=108$

Cupcakes made by B = $\frac{540}{9}=60$

Person A made 108 cupcakes and person B made 60 cupcakes.

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3 years ago

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3 years ago

A small plane took 3 hours to fly 960 km from Ottawa to Halifax with a tail wind. On the return trip, flying into the wind, the

Rina8888 [55]

Answer:

Wind speed: $\rm 40\; km \cdot h^{-1}$ .
Speed of the plane in still air: $\rm 320\; km \cdot h^{-1}$ .

Step-by-step explanation:

This problem involves two unknowns:

wind speed, and
speed of the plane in still air.

Let the speed of the wind be $x \rm \; km \cdot h^{-1}$ , and the speed of the plane in still air be $y\rm \; km \cdot h^{-1}$ . It takes at least two equations to find the exact solutions to a system of two variables.

Information in this question gives two equations:

It takes the plane three hours to travel $\rm 960\; km$ from Ottawa to with a tail wind (that is: at a ground speed of $x + y$ .)
It takes the plane four hours to travel $\rm 960\; km$ from Halifax back to Ottawa while flying into the wind (that is: at a ground speed of $-x + y$ .)

Create a two-by-two system out of these two equations:

$\left\{ \begin{aligned}&3(x + y) = 960 && (1) \\ &4(-x + y) = 960 && (2) \end{aligned}\right.$ .

There can be many ways to solve this system. The approach below avoids multiplying large numbers as much as possible.

Note that this system is equivalent to

$\left\{ \begin{aligned}&4 \times 3 (x + y) = 4\times960 && 4 \times (1) \\ &3\times 4(-x + y) = 3\times 960 && 3 \times (2) \end{aligned}\right.$ .

$\left\{ \begin{aligned}&12 x + 12y = 4\times960 && 4 \times (1) \\ &- 12x + 12y = 3\times 960 && 3 \times (2) \end{aligned}\right.$ .

Either adding or subtracting the two equations will eliminate one of the variables. However, subtracting them gives only $1 \times 960$ on the right-hand side. In comparison, adding them will give $7 \times 960$ , which is much more complex to evaluate. Subtracting the second equation ( $3 \times (2)$ ) from the first ( $4 \times (1)$ ) will give the equation

$(12 - (-12) x = 1 \times 960$ .

$24 x = 960$ .

$x = 40$ .

Substitute $x$ back into either equation $(1)$ or $(2)$ of the original system. Solve for $y$ to obtain $y = 320$ .

In other words,

Wind speed: $\rm 40\; km \cdot h^{-1}$ .
Speed of the plane in still air: $\rm 320\; km \cdot h^{-1}$ .

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3 years ago

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