Can an amount be increased by more than 100%. If you explain I’ll make you brainliest.

Mathematics

2 answers:

sergejj [24]3 years ago

6 0

Answer:

It's impossible to give more than 100%, they'll say. That's what percent means. But of course percentages greater than 100 are possible. It all depends on what your baseline is — x percent of what.

Step-by-step explanation:

DiKsa [7]3 years ago

5 0

Answer:

Step-by-step explanation:

Yes, it can.

A 100% increase means it's the original amount plus the original amount (2x). 100+100=200

A 150% increase would be the original amount plus 1.5 times the original amount. 100+150=250

You might be interested in

e mean IQ for 100 residents of the local apartment complex is 110. What is the best point estimate for the mean IQ for all resid

KengaRu [80]

Answer:

110 IQ too!

Step-by-step explanation:

For statistical problems, especially for ones with large enough sample sizes like this one (n = sample size = 100), the means for various samples is usually equal to the population mean.

It is the standard deviation of sample that differs from the standard deviation of the population through

σₓ = σ/√n

But, The mean of the sample = The population mean

μₓ = μ = 110 IQ

8 0

3 years ago

HELP ME MATH ILL GIVE YOU BRAINLIEST

s344n2d4d5 [400]

Answer:

zeros: x = -3, -1, +2.
end behavior: as x approaches -∞, f(x) approaches -∞.

Step-by-step explanation:

I like to use a graphing calculator for finding the zeros of higher order polynomials. The attachment shows them to be at x = -3, -1, +2.

The zeros can also be found by trial and error, trying the choices offered by the rational root theorem: ±1, ±2, ±3, ±6. It is easiest to try ±1. Doing so shows that -1 is a root, and the residual quadratic is ...

x² +x -6

which factors as (x -2)(x +3), so telling you the remaining roots are -3 and +2.

___

For any odd-degree polynomial with a positive leading coefficient, the sign of the function will match the sign of x when the magnitude of x gets large. Thus as x approaches negative infinity, so does f(x).