Identify the two whole numbers between which the product lies 6*4/5

Solve and in standard form please. you must add. this is from khan academy

Sindrei [870]

We are asked to sum the two trinomials $(-4f^3-5f+16)+(4f^2-f+9)$ . We can start by using the distributive property to remove the parenthesis, giving us $-4f^3-5f+16+4f^2-f+9$ . Next, when we combine like terms, we get $\boxed{-4f^3+4f^2-6f+25}$ . Hope this helped!

8 0

3 years ago

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6x - 8y - 5х +3y. How can you answer this

insens350 [35]

Step-by-step Solution

Problem to solve: 6x−8y−5х+3y

solving method: simplifying

Combining like terms -8yand 3y

6x−5y−5

Final Answer:

6x−5y−5

8 0

3 years ago

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Can anybody please tell me the answers to these questions because I'm really confused

dimulka [17.4K]

3. 3/10 and 5/10 4. 12/20 and 15/20 5.4/8 and 7/8 6. 8/12 and 5/12 7. 3/12 and 2/12. 8. 1/2>2/5 9.1/2 = 3/6 10. 3/4<5/6 11. 6/10=3/5

8 0

3 years ago

Can someone please solve this ASAP?

olga2289 [7]

The ladder and the wall form a right triangle.
The length is the hypotenuse = 15.
The base = 2*(15/5) = 2*3 = 6

Use pythagorean theorem to find other side of triangle:
$a^2 + b^2 = c^2 \\ \\ a^2 + 6^2 = 15^2 \\ \\ a^2 + 36 = 225 \\ \\ a^2 = 189 \\ \\ a = \sqrt{189} = 3 \sqrt{21}$

8 0

3 years ago

A sample of eight workers in a clothing manufacturing company gave the following figures for the amount of time(in minutes) need

lilavasa [31]

Answer:

$10.108 < \mu < 13.892$

Step-by-step explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

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".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

We have the following distribution for the random variable:

$X \sim N(\mu , \sigma=0.45)$

And by the central theorem we know that the distribution for the sample mean is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

2) Confidence interval

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the mean and the sample deviation we can use the following formulas:

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$ (2)

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$ (3)

The mean calculated for this case is $\bar X=12$

The sample deviation calculated $s=2.268$

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=8-1=7$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.025,7)".And we see that $t_{\alpha/2}=2.36$

Now we have everything in order to replace into formula (1):

$12-2.36\frac{2.268}{\sqrt{8}}=10.108$

$12+2.36\frac{2.268}{\sqrt{8}}=13.892$

So on this case the 95% confidence interval would be given by (10.108;13.892)

$10.108 < \mu < 13.892$

7 0

3 years ago