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AysviL [449]
3 years ago
9

If 5x+6-19=5 what is x

Mathematics
2 answers:
Troyanec [42]3 years ago
5 0

Answer:

<em>x </em>equals 3.6.

Step-by-step explanation:

A way to solve this equation is to preform the inverse operations on both sides.

1. Add 19 to -19 and 5: 5<em>x </em>+ 6 - 19 + 19 = 5 + 19  ---> 5x + 6 = 24

2. Subtract 6 from 6 and 5: 5x + 6 - 6 = 24 - 6 ---> 5x = 18

3. Divide 5x and 18 by 5: 5x / 5 = 18 / 5 ---> x = 3.6

4. 3.6 is the answer.

To prove it, substitute x with 3.6 in the equation: 5 · 3.6 + 6 - 19 = 5

I hope this made sense and helped you a lot! :)

Effectus [21]3 years ago
5 0
Answer:
x equals to 3.6
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t=\frac{201-190}{\frac{10}{\sqrt{25}}}=5.5    

p_v =P(t_{(24)}>5.5)=0.00000589  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the true mean is higher than 190

Step-by-step explanation:

Data given and notation  

\bar X=201 represent the mean

s=10 represent the sample standard deviation for the sample  

n=25 sample size  

\mu_o =190 represent the value that we want to test

\alpha=0.05 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is higher than 190, the system of hypothesis would be:  

Null hypothesis:\mu \leq 190  

Alternative hypothesis:\mu > 190  

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

t=\frac{201-190}{\frac{10}{\sqrt{25}}}=5.5    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=25-1=24  

Since is a one side test the p value would be:  

p_v =P(t_{(24)}>5.5)=0.00000589  

Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the true mean is higher than 190

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