If you have more expenses than income, which is NOT a good way to balance your budget?

Mathematics

1 answer:

Alex Ar [27]3 years ago

5 0

I think that increasing ur expenses won’t help balance ur budget
Hope that helps:)

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Find the TWO integers whos product is -12 and whose sum is 1<br>m

svetoff [14.1K]

Answer:

$\rm Numbers = 4 \ and \ -3.$

Step-by-step explanation:

Given :-

The sum of two numbers is 1 .

The product of the nos . is 12 .

And we need to find out the numbers. So let us take ,

First number be x

Second number be 1-x .

According to first condition :-

$\rm\implies 1st \ number * 2nd \ number= -12\\\\\rm\implies x(1-x)=-12\\\\\rm\implies x - x^2=-12\\\\\rm\implies x^2-x-12=0\\\\\rm\implies x^2-4x+3x-12=0\\\\\rm\implies x(x-4)+3(x-4)=0\\\\\rm\implies (x-4)(x+3)=0\\\\\rm\implies\boxed{\red{\rm x = 4 , -3 }}$

Hence the numbers are 4 and -3.

5 0

3 years ago

Read 2 more answers

The mean number of words per minute (WPM) read by sixth graders is 97 with a standard deviation of 19 WPM. If 75 sixth graders a

andrey2020 [161]

Answer:

0.0174 = 1.74% probability that the sample mean would be greater than 101.63 WPM

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ ;