Suppose that salaries for recent graduates of one university have a mean of $25,700 with a standard deviation of $1350. using chebyshev's theorem, state the range in which at least 75% of the data will reside. do not round your answers.

75% of the data will reside in the range 23000 to 28400.

Step-by-step explanation :

To find the range of values :

We need to find the values that deviate from the mean. Since we want at least 75% of the data to reside between the range therefore we have,

Solving this, we would get k = 2 which shows the value one needs to find lies outside the range.

Range is given by : mean +/- (z score) × (value of a standard deviation)  

⇒ Range : 25700 +/- 2 × 1350

⇒ Range : (25700 - 2700) to (25700 + 2700)

Hence, 75% of the data will reside in the range 23000 to 28400.

there are many solutions.

step-by-step explanation:

step 1: simplify both sides of the equation.

4x+6=4x^2+10x−4

step 2: subtract 4x^2+10x-4 from both sides.

4x+6−(4x^2+10x−4)=4x^2+10x−4−(4x^2+10x−4)

−4x^2−6x+10=0

step 3: factor left side of equation.

−2(2x+5)(x−1)=0

step 4: set factors equal to 0.

x=-5/2 or x=1

step-by-step explanation: