Ask question

Ask question

All categories

3 years ago

6

How do I solve?

TexFormula1" title="2 = \frac{x(x-1)(x-3)^{2} (x+1)}{x}" alt="2 = \frac{x(x-1)(x-3)^{2} (x+1)}{x}" align="absmiddle" class="latex-formula">

2 answers:

Rus_ich [418]3 years ago

8 0

Answer:

x = 5

Step-by-step explanation:

x(x - 1)(x - 3)^2(x + 1)

x - x + x - 9 + x + 1

x - 9 + x + 1

2x - 8 = 2

<u> +8 +8</u>

2x = 10

<em><u>x = 5</u></em>

Alexxandr [17]3 years ago

6 0

Answer:

x=5

Step-by-step explanation:

You might be interested in

On a hiking trip,LaShana notes that she hikes about 12 kilometers every 4 hours .If she continues at this rate,use a ratio table

Evgesh-ka [11]

12 : 4 = 3 km her <span>speed per hour
3 * 6 = 18 km </span><span>she could hike in 6 hours</span>

4 0

4 years ago

What is the product (9t-4)(-9-4)

Scrat [10]

Answer:

-81t^2 +16

Step-by-step explanation:

(9t-4)(-9t-4)

FOIL

first: -9t*9t = -81t^2

outer : -4*9t = -36t

inner : -4 *-9t = 46t

last = -4*-4 = 16

Add them together

-81t^2 -36t+36t+16

-81t^2 +16

6 0

3 years ago

Find the value of x in each case.

prisoha [69]

It’s 282 because if you add all those rhe divide you get the answer

3 0

3 years ago

A random sample of 250 students at a university finds that these students take a mean of 15.3 credit hours per quarter with a st

SpyIntel [72]

Answer: $15.3\pm0.198$

OR

(15.102, 15.498)

Step-by-step explanation:

The formula to find the confidence interval $(\mu)$ is given by :-

$\overline{x}\pm z_{\alpha/2, df}\dfrac{s}{\sqrt{n}}$

, where n is the sample size

$s$ = sample standard deviation.

$\overline{x}$ = Sample mean

$z_{\alpha/2}$ = Two tailed z-value for significance level of $\alpha$ .

Given : Confidence level = 95% = 0.95

Significance level = $\alpha=1-0.95=0.05$

$s= 1.6$

$\overline{x}=15.3$

sample size : n= 250 , which is extremely large ( than n=30) .

So we assume sample standard deviation is the population standard deviation.

thus , $\sigma=1.6$

By standard normal distribution table ,

Two tailed z-value for Significance level of 0.05 :

$z_{\alpha/2}=z_{0.025}=1.96$

Then, the 95% confidence interval for the mean credit hours taken by a student each quarter :-

$15.3\pm (1.96)\dfrac{1.6}{\sqrt{250}}\\\\ =15.3\pm 0.19833\\\\=\approx15.3\pm0.198\\\\=(15.3-0.198,\ 15.3+0.198)=(15.102,\ 15.498)$

Hence, the mean credit hours taken by a student each quarter using a 95% confidence interval. =(15.102, 15.498)

5 0

3 years ago

If the point A is translated 5 units up and 10 units to the left, find the location of the resulting point A' ? PLEASE HELP!!

Ugo [173]

Answer:

6, -5

Step-by-step explanation:

8 0

3 years ago