Animals: The label on a bottle of pet vitamins lists dosage guidelines. What dosage would you give to each of these animals? (a)

ANSWER:

(-5, 0.5) serves as a solution for the equations x – 4y = –7, 0.2x + 2y = 0.

SOLUTION:

We start with the two equations: x – 4y = -7 → (1)

And 0.2x + 2y = 0 → (2)

We will verify if (-5, 0.5) is a solution.

To do this, we will resolve both equations.

Let’s multiply equation (2) by 2 to facilitate the cancellation of y terms.

This transforms equation (2) into:

0.4x + 4y = 0 → (3)

Now, combine equations (1) and (3), leading to:

1.4x + 0 = -7

Next, input the x value into (2)

0.2(-5) + 2y = 0

-1 + 2y = 0

2y = 1

y = 0.5

Thus, the result for the equations is (-5, 0.5).

Therefore, (-5, 0.5) is the solution to the equations.

Answer:

We require a total of 144 birthday candles for the 12 parties.

Step-by-step explanation:

As a caterer focused on children’s birthday celebrations, you have 12 events to cater to in the upcoming week, each requiring 2 cakes. Each cake will be adorned with 6 candles.

Thus, for all 12 birthday gatherings, you need a total of =

Further, since every cake holds 6 candles, the overall number of candles for the 24 cakes amounts to =

Answer:

(a1) The chance that the temperature rises by less than 20°C is 0.667.

(a2) The probability of the temperature increase falling between 20°C and 22°C is 0.133.

(b) The likelihood that the temperature increase could be hazardous at any moment is 0.467.

(c) The anticipated value of the temperature rise is 17.5°C.

Step-by-step explanation:

Let X denote the temperature increase.

The random variable X is distributed uniformly over the interval [10°C, 25°C].

The probability density function for X is shown here:

(a1)

The probability of the temperature increase being under 20°C can be calculated as follows:

Consequently, the chance that the temperature increase will be below 20°C is 0.667.

(a2)

The probability of the temperature rise being in the range from 20°C to 22°C is computed as follows:

This leads to the probability of the temperature increase being between 20°C and 22°C being 0.133.

(b)

To find the probability that the increase in temperature could be dangerous, we calculate:

This results in a probability of 0.467 that the temperature rise is potentially dangerous at any time.

(c)

The expected value of the uniform random variable X is determined as follows:

The expected value for the temperature increase computes to 17.5°C.