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3 years ago

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Stefan sells Jin a bicycle for 138 $ and a helmet for 16 $. The total cost for Jin is 140% of what Stefan spent originally to

buy the bike and helmet. How much did Stefan spend originally? How much money did he make by selling the bicycle and helmet to Jin?

1 answer:

kykrilka [37]3 years ago

7 0

Answer:

its 3

Step-by-etep explanation:

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Parallelogram ABCD has the angle measures shown. Can you conclude that it is a rhombus, a rectangle, or a square? Explain.

Nikitich [7]

<span>We can analyze the four optons. 1) Option A. A parallelogram with all four angles of the same measure can be either a square or a rectangle, then this option is not valid. 2) Optrion B. gives not information. 3) A rhombus (a diamond) is a parallelogram with four congruent side (square is a specific case of rhmbus but not all rhombus are squares), and it is enouh to say that one diagonal bisects two interior angles, to conclude that it is a rhombus. 4) If a diagonal creates congruent angles, but you do not know what happens with the opposed angle, you cannot conclude that the parallelogram is a rectangle; it could be a trapezoid with one side perpendicular to the parallel sides. By t his analysis, the answer is option C.</span>

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3 years ago

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EMERGENCY!!!!<br><br> 15-26<br><br> Please slove all of them it’s due tomorrow

DanielleElmas [232]

I don’t have the time to do this, but there is a app called Socratic where you can take a pic of your homework and it searches everything on google.

That’s how I got Brainly

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3 years ago

Flow meters are installed in urban sewer systems to measure the flows through the pipes. In dry weatherconditions (no rain) the

ziro4ka [17]

Answer:

a) $\frac{(8)(30.23)^2}{20.09} \leq \sigma^2 \leq \frac{(8)(30.23)^2}{1.65}$

$363.90 \leq \sigma^2 \leq 4430.80$

Now we just take square root on both sides of the interval and we got:

$19.08 \leq \sigma \leq 66.56$

b) For this case we are 98% confidence that the true deviation for the population of interest is between 19.08 and 66.56

Step-by-step explanation:

423.6, 487.3, 453.2, 402.9, 483.0, 477.7, 442.3, 418.4, 459.0

Part a

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

On this case we need to find the sample standard deviation with the following formula:

$s=sqrt{\frac{\sum_{i=1}^8 (x_i -\bar x)^2}{n-1}}
And in order to find the sample mean we just need to use this formula:
[tex]\bar x =\frac{\sum_{i=1}^n x_i}{n}$

The sample deviation for this case is s=30.23

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

$df=n-1=9-1=8$

The Confidence interval is 0.98 or 98%, the value of $\alpha=0.02$ and $\alpha/2 =0.01$ , and the critical values are:

$\chi^2_{\alpha/2}=20.09$

$\chi^2_{1- \alpha/2}=1.65$

And replacing into the formula for the interval we got:

$\frac{(8)(30.23)^2}{20.09} \leq \sigma^2 \leq \frac{(8)(30.23)^2}{1.65}$

$363.90 \leq \sigma^2 \leq 4430.80$

Now we just take square root on both sides of the interval and we got:

$19.08 \leq \sigma \leq 66.56$

Part b

For this case we are 98% confidence that the true deviation for the population of interest is between 19.08 and 66.56

4 0

3 years ago

The polynomial x^2+3x-1 is a factor of x^4+3x^3-2x^2-3x+1 true or false?

pashok25 [27]

Answer:

Yes, it is true that $x^2+3x-1$ is a factor of $x^4+3x^3-2x^2-3x+1$ .

Step-by-step explanation:

Let us try to factorize $x^4+3x^3-2x^2-3x+1$

$x^4+3x^3-2x^2-3x+1\\\Rightarrow x^4-2x^2+1-3x+3x^3$

Let us try to make a whole square of the given terms:

$\Rightarrow (x^2)^2-2\times x^2 \times 1+1^2+3x^3-3x\\\Rightarrow (x^2-1)^2+3x^3-3x\\$

--------------

Formula used above:

$a^{2} -2 \times a \times b +b^{2} = (a-b)^2$

In the above equation, we had $a = x, b = 1$ .

--------------

Further solving the above equation, taking $3x$ common out of $3x^3-3x$

$\Rightarrow (x^2-1)^2+3x(x^2-1)\\$

Taking $(x^{2} -1)$ common out of the above term:

$\Rightarrow (x^2-1)((x^2-1)+3x)\\\Rightarrow (x^2-1)(x^2+3x-1)$

So, the two factors are $(x^2-1)\ and\ (x^2+3x-1)$ .

$\therefore$ The statement that $x^2+3x-1$ is a factor of $x^4+3x^3-2x^2-3x+1$ is <em>True.</em>

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2 years ago

What do I do on this?

stiks02 [169]

All you do is measure it with a ruler and for every 1 cm you get multiply it by 8.5ft.
so for example if you got 6cm then you would multiply it by 8.5ft and your answer would be 51ft.

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3 years ago