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3 years ago

Linear equations- please help me −5(5x+3)−5x+1=−44

1 answer:

5 0

The answer is x=1
Add the numbers
Combine like terms
Add
to both sides of the equation

Simplify

Divide both sides of the equation by the same term

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5x−2.5+6x−3= ___ (2x−1)

LiRa [457]

Answer:

5.5

Step-by-step explanation:

So, this is a simple factoring problem. The answer is 5.5(2x-1). So, heres how we get that answer.

So, factoring is basically breaking down the problem into smaller parts. So once we simplify the problem, we can factor it.

Simplify by combining like terms: 5x+6x= 11x and -2.5-3=-5.5

Now we have 11-5.5=?

so, we need to have (2x-1) in our answer because the problem asks for that in the answer.

We need to find a number that satisfies the equation and answer

so, 5.5(2x) is 11x, and 5.5(-1) is 5.5.

5.5 satisfies the equation, therefore is the answer

4 0

4 years ago

Read 2 more answers

PLZ HELP ASAP MATH ISN'T MY THING !!!<br>@lastbenchstudent

dezoksy [38]

your Answer is Option A

6 0

3 years ago

How to write 2.31818 as a mixed fraction

Elenna [48]

Answer:

$2\frac{15909}{5000}$

Step-by-step explanation:

To make $2.31818$ into a fraction, you must first move it to the numerator and make 1 the denominator.

$\frac{2.31818}{1}$

And then you multiply it by $10,000$ or you can say $10^5$

$\frac{2.31818}{1}*\frac{100000}{100000}$

After you multiply you divide the numerator and denominator by 2.

$\frac{231818/2}{100000/2}$

And now this equals:

$2\frac{15909}{5000}$

Hope this helps!

5 0

3 years ago

Select the works written by Augustine of Hippo.

Alex73 [517]

Answer:

Consolation of Philosopy..

5 0

4 years ago

Test the claim that the mean GPA of night students is larger than 2 at the .025 significance level. The null and alternative hyp

exis [7]

Answer:

$H_0: \, \mu = 2$ .

$H_1:\, \mu > 2$ .

Test statistics: $z \approx 2.582$ .

Critical value: $z_{1 - 0.025} \approx 1.960$ .

Conclusion: reject the null hypothesis.

Step-by-step explanation:

The claim is that the mean $\mu$ is greater than $2$ . This claim should be reflected in the alternative hypothesis:

$H_1:\, \mu > 2$ .

The corresponding null hypothesis would be:

$H_0:\, \mu = 2$ .

In this setup, the null hypothesis $H_0:\, \mu = 2$ suggests that $\mu_0 = 2$ should be the true population mean of GPA.

However, the alternative hypothesis $H_1:\, \mu > 2$ does not agree; this hypothesis suggests that the real population mean should be greater than $\mu_0= 2$ .

One way to test this pair of hypotheses is to sample the population. Assume that the population mean is indeed $\mu_0 = 2$ (i.e., the null hypothesis is true.) How likely would the sample (sample mean $\overline{X} = 2.02$ with sample standard deviation $s = 0.06$ ) be observed in this hypothetical population?

Let $\sigma$ denote the population standard deviation.

Given the large sample size $n = 60$ , the population standard deviation should be approximately equal to that of the sample:

$\sigma \approx s = 0.06$ .

Also because of the large sample size, the central limit theorem implies that $Z= \displaystyle \frac{\overline{X} - \mu_0}{\sigma / \sqrt{n}}$ should be close to a standard normal random variable. Use a $Z$ -test.

Given the observation of $\overline{X} = 2.02$ with sample standard deviation $s = 0.06$ :

$\begin{aligned}z_\text{observed}&= \frac{\overline{X} - \mu_0}{\sigma / \sqrt{n}} \\ &\approx \frac{\overline{X} - \mu_0}{s / \sqrt{n}} = \frac{2.02 - 2}{0.06 / \sqrt{60}} \approx 2.582\end{aligned}$ .

Because the alternative hypothesis suggests that the population mean is greater than $\mu_0 = 2$ , the null hypothesis should be rejected only if the sample mean is too big- not too small. Apply a one-sided right-tailed $z$ -test. The question requested a significant level of $0.025$ . Therefore, the critical value $z_{1 - 0.025}$ should ensure that $P( Z > z_{1 - 0.025}) = 0.025$ .

Look up an inverse $Z$ table. The $z_{1 - 0.025}$ that meets this requirement is $z_{1 - 0.025} \approx 1.960$ .

The $z$ -value observed from the sample is $z_\text{observed}\approx 2.582$ , which is greater than the critical value. In other words, the deviation of the sample from the mean in the null hypothesis is sufficient large, such that the null hypothesis needs to be rejected at this $0.025$ confidence level in favor of the alternative hypothesis.

3 0

3 years ago