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2 years ago

What is the solution to the inequality 7x + 2 < -3x + 1?

Mathematics

1 answer:

anastassius [24]2 years ago

5 0

Answer:

$\Large\boxed{\sf{x < -\dfrac{1}{10} }}$

Step-by-step explanation:

In order to find the solutions, you must isolate the inequality on one side of the equation.

7x+2<-3x+1

First, you subtract by 2 from both sides.

7x+2-2<-3x+1-2

Solve.

Subtract the numbers from left to right.

7x<-3x-1

Add by 3x from both sides.

7x+3x<-3x-1+3x

Solve.

7x+3x=10x

10x<-1

Then, you divide by 10 from both sides.

10x/10<-1/10

Solve.

Divide these numbers goes from left to right.

$\Longrightarrow: \boxed{\sf{x < -\dfrac{1}{10} }}$

Therefore, the final answer is x<-1/10.

I hope this helps, let me know if you have any questions.

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It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minute

Molodets [167]

Answer:

10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

99.55% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

Normal probability distribution

Problems of normally distributed samples can be 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.

Mildly obese

Normally distributed with mean 375 minutes and standard deviation 68 minutes. So $\mu = 375, \sigma = 68$

What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes?

So $n = 6, s = \frac{68}{\sqrt{6}} = 27.76$

This probability is 1 subtracted by the pvalue of Z when X = 410.

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

$Z = \frac{410 - 375}{27.76}$

$Z = 1.26$

$Z = 1.26$ has a pvalue of 0.8962.

So there is a 1-0.8962 = 0.1038 = 10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

Lean

Normally distributed with mean 522 minutes and standard deviation 106 minutes. So $\mu = 522, \sigma = 106$