a desert has both fruit and yogurt inside altogether the mass of the desert is 185g the ratio of the mass to fruit to the mass o

We initiate with:

|X+6|- 12 < 13

We can then add 12 to both sides:

|X+6| < 13 + 12 = 25

|X+6| < 25

This leads us to form two separate equations:

if x > -6, consequently |X+6|  = x + 6

        (x + 6) < 25

         x < 25 - 6

          x <  19

If x < -6, so |X+6|  = -(x + 6)

         -(x + 6) < 25

          -x < 31

            x > -31

From these equations, we find:

-31 < x < 19

A meticulous method to create this would involve partitioning the spinner into 100 segments and coloring 5 of those. This results in a probability of 5/100 = 0.05 = 5%, indicating that each of the 5 segments has a 5% likelihood of being landed upon.

A more efficient approach is to split it into 20 segments instead. This works because simplifying 5/100 yields 1/20 (by dividing both the numerator and denominator by 5). Thus, instead of coloring in 5 segments, Mark should just color 1. If you check this with a calculator, you’ll see that 1/20 = 0.05 = 5% too.

The second choice is more efficient, but a spinner with 20 pieces is still somewhat cumbersome. It may be more practical for Mark to draw random numbers from a hat or use a computer to generate random numbers.

Answer:

The distance equals 816 km.

Step-by-step explanation:

5/6 multiplied by 4/1 equals 20/6.

Dividing 20/6 by 2 results in 3 1/3.

This triangle is both acute and equilateral.
It consists of three angles that are all acute, as well as having equal sides and angles.

Answer:

The recorded temperature is -0.675ºC.

Detailed explanation:

To tackle problems involving normally distributed samples, the z-score formula can be utilized.

In a distribution with mean and standard deviation , the z-score for a specific measure X is calculated as follows:

The Z-score indicates how many standard deviations a given measure deviates from the mean. Once the Z-score is determined, we refer to the z-score table to obtain the corresponding p-value. This p-value represents the likelihood that the measure's value is less than X, thereby indicating the percentile of X. By taking 1 minus the p-value, we find the probability that the measure's value exceeds X.

For this scenario, we know that:

Assuming the thermometer readings follow a normal distribution with a mean of 0◦ and a standard deviation of 1.00◦C, this leads us to

We need to determine P25, which is the 25th percentile.

This represents the value of X corresponding to Z with a p-value of 0.25, thus we utilize , applicable between and .

The recorded temperature is -0.675ºC.