How do you solve 3|4x+2|=6

kirill115 [55]

The answer is 3 because when you give the fractions a common denominator of 54, you then add the negative 45/54 and the positive 48/ 54 and you get 3/54, then you divide the 3/54 by 9/54 and you get 9/3, which is the same thing as 3.

So the answer is 3

7 0

2 years ago

Simplify the expression8(5 plus w)

inna [77]

8(5 plus w)

distribute

8*5 + 8*w

40 + 8w

8 0

3 years ago

Please help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

erma4kov [3.2K]

Step-by-step explanation:

so option 1 is rightttt

5 0

2 years ago

!!WILL MARK BRAINLIEST!! <br><br> plz help this is due today

meriva

Answer:

The answer is A

Step-by-step explanation:

3 0

2 years ago

A random sample of n = 40 observations from a quantitative population produced a mean x = 2.2 and a standard deviation s = 0.29.

zmey [24]

Answer:

$t=\frac{2.2-2.1}{\frac{0.29}{\sqrt{40}}}=2.18$

$df=n-1=40-1=39$

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

$p_v =P(t_{(39)}>2.18)=0.0177$

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 then the population mes seems to be higher than 2.1 at 5% of significance

Step-by-step explanation:

Data given and notation

$\bar X=2.2$ represent the sample mean

$s=0.29$ represent the sample standard deviation

$n=40$ sample size

$\mu_o =2.1$ 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 2,1, the system of hypothesis would be:

Null hypothesis: $\mu \leq 2.1$

Alternative hypothesis: $\mu > 2.1$

If we analyze the size for the sample is > 30 but 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:

$t=\frac{2.2-2.1}{\frac{0.29}{\sqrt{40}}}=2.18$

P-value

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

$df=n-1=40-1=39$

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

$p_v =P(t_{(39)}>2.18)=0.0177$

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 then the population mes seems to be higher than 2.1 at 5% of significance

7 0

3 years ago