What is true about the solution of StartFraction x squared Over 2 x minus 6 EndFraction = StartFraction 9 Over 6 x minus 18 EndF

Réponse:

Le domaine est l'ensemble des nombres réels

y ≤ 16

Explication étape par étape:

Le domaine est l'ensemble des valeurs x que le graphe couvre. Il n'y a aucune restriction sur les valeurs x. Le domaine est l'ensemble des nombres réels.

La plage est l'ensemble des valeurs y que le graphe couvre. Les valeurs y atteignent au maximum 16 au sommet de la parabole. Ainsi, la plage est y ≤ 16.

Answer:

Step-by-step explanation:

I don't know

Detailed explanation:

Thus,

100% plus 8% equals 108%

108% equals 1215

1% corresponds to 11.25

Therefore, 100% amounts to 1125 rs

Answers:

A. Ms. Dobson’s class exhibits a narrower score range;

B. The district demonstrates a higher interquartile range; and

C. The score of fifteen is an outlier for the district.

Explanation:

The score range in Ms. Dobson's class spans from a low of 40 to a high of 95, which results in a range of 95-40 = 55.  In contrast, the district's scores range from 15 to 100, yielding a range of 100-15 = 85.  Thus, Ms. Dobson's class possesses a smaller score range.

The interquartile range (IQR) is attained by subtracting Q1, the first quartile, from Q3, the third quartile.

For Ms. Dobson's class, Q1 equals 70 and Q3 equals 80, leading to an IQR of 80-70 = 10.  In the district, Q1 is 55 and Q3 is 75, resulting in an IQR of 75-55 = 20.  Thus, the district has a more extensive interquartile range.

An outlier is defined as any score that falls below 1.5 times the interquartile range from Q1 or above 1.5 times the interquartile range from Q3.

For the district, Q1 is 55 and the IQR is 20; thus, an outlier would be below

55-1.5(20) = 55-30 = 25.  Since 15 is under 25, it qualifies as an outlier.

For the district, Q3 is 75 and the IQR is 20; thus, an outlier would be above 75+1.5(20) = 75+30 = 105; no scores within the district reach this level, and 100 does not consider an outlier for this set.

Overall, Ms. Dobson's class achieved higher scores than the district, despite a few higher scores within the district, as most of Ms. Dobson’s class scores were notably clustered higher.

Answer: The other two observations are 97 and 107.

Explanation:

We know that

Mean = 100

Mode = 98

Range = 10

And from the formula,

Range = Highest - Lowest.

Let’s set the highest observation as x and the lowest as y.

Thus, we have the equation x - y = 10 (equation 1).

The observations can be represented as:

x, 98, 98, y.

Using the mean formula yields:

Mean = .

This means our second equation is:

x + y = 204.

By applying the elimination method to solve these linear equations, we find:

x = 97.

and

.

Therefore, the other two observations are 97 and 107.