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3 years ago

Solve x in 3x+2+5x=4(2x-1)+x Step by step explanation please ❤️

Mathematics

1 answer:

Aleks [24]3 years ago

5 0

Answer:

NO SOLUTION

Step-by-step explanation:

Start by combining like terms: combine all the x terms and then all the constants:

8x + 2 = 8x - 4

Subtracting 8x from both sides results in 2 = -4, which is never true. Therefore, this equation has NO SOLUTION.

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Geomtry homework please help

ikadub [295]

Answer:

Your answer is 116

Hope that this is helpful. Tap the crown button, Mark my answer as............. BRAINLIST......... Like & Follow me to get more answer.

6 0

3 years ago

pleaseee help context is down below $3 is the entrance fee while $.50 is for each game played that's enough evidence to solve C(

BabaBlast [244]

C= Cost
g= # of times games played

C = 0.50g + 3

C(0) = 0.50(0) + 3
C(0) = 0 + 3
C(0) = 3

This means that no games have been played yet and the entrance fee is $3

5 0

2 years ago

Read 2 more answers

78 divided by 190 steps

Goryan [66]

78 x 100 = 7800

How many times does 190 go into 780? 
≈ 4....190 x 4 = 760 

780 - 760 = 20...bring down the extra 0 to make it 200. 

How many times does 190 go into 200? 
≈ 1...subtract 190 from 200 to get a remainder of 10 

190 ÷ 7800 ≈ 41

Hope I Helped You!!! :-)

Have A Good Day!!!

5 0

3 years ago

ANSWER QUICK PLEASE ANSWER part 2 I WILL MARK BRAINLIEST (:

FrozenT [24]

The answer is A, not sure about the expression though. Sorry

4 0

3 years ago

The local oil changing business is very busy on Saturday mornings and is considering expanding. A national study of similar busi

eimsori [14]

Answer:

$t=\frac{4.2-3.6}{\frac{1.4}{\sqrt{16}}}=1.714$

Reject the null hypothesis if the observed "t" value is less than -2.131 or higher than 2.131

Rejection Zone: $t_{calculated}$ or $t_{calculated}>2.131$

In our case since our calculated value is not on the rejection zone we don't have enough evidence to reject the null hypothesis at 5% of significance.

Step-by-step explanation:

Previous concepts and data given

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=4.2$ represent the sample mean

$s=1.4$ represent the sample standard deviation

n=16 represent the sample selected

$\alpha=0.05$ significance level

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is different from 3.6, the system of hypothesis would be:

Null hypothesis: $\mu = 3.6$

Alternative hypothesis: $\mu \neq 3.6$

If we analyze the size for the sample is < 30 and we 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:

$t=\frac{4.2-3.6}{\frac{1.4}{\sqrt{16}}}=1.714$

Critical values

On this case since we have a bilateral test we need to critical values. We need to use the t distribution with $df=n-1=16-1=15$ degrees of freedom. The value for $\alpha=0.05$ and $\alpha/2=0.025$ so we need to find on the t distribution with 15 degrees of freedom two values that accumulates 0.025 of the ara on each tail. We can use the following excel codes:

"=T.INV(0.025,15)" "=T.INV(1-0.025,15)"

And we got $t_{crit}=\pm 2.131$

So the decision on this case would be:

Reject the null hypothesis if the observed "t" value is less than -2.131 or higher than 2.131

Rejection Zone: $t_{calculated}$ or $t_{calculated}>2.131$

Conclusion

In our case since our calculated value is not on the rejection zone we don't have enough evidence to reject the null hypothesis at 5% of significance.

3 0

3 years ago