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2 years ago

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Sergio walked to and from school on 3/5 of the days in one month. He got a ride to school on 7/8 of the remaining days. On the o

ne remaining school day that month he stayed home with a cold. How many school days were there that month?

2 answers:

Talja [164]2 years ago

6 0

Possibly 14 because 5+8 plus the one day he was out

horrorfan [7]2 years ago

4 0

14 days ! Hope it helped

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Answer: There is a probability of 0.05 that there is neither truck is available.

Step-by-step explanation:

Since we have given that

Probability that the first truck is available = 0.75

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Probability that both trucks are available = 0.30

So, probability that either first truck or second truck is available is given by

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We need to find the probability that neither truck is available.

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Recall the equation for a circle with center (h, k) and radius r. At what point in the first quadrant does the line with equatio

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Answer:

The point of intersection is:

$\displaystyle \left(\frac{6\sqrt{5}}{5}, \frac{3\sqrt{5}}{5}+5\right)\approx \left(2.68, 6.34)$

Step-by-step explanation:

We want to find the point in QI at which the line with the equation:

$y=0.5x+5$

Intersect a circle with a radius of 3 and a center of (0, 5).

First, write the equation of a circle. The equation for a circle is given by:

$(x-h)^2+(y-k)^2=r^2$

Where (<em>h, k</em>) is the center and <em>r</em> is the radius.

Since our center is (0, 5), <em>h</em> = 0 and <em>k</em> = 5. The radius is 3. So, <em>r</em> = 3. Substitute:

$(x-0)^2+(y-5)^2=(3)^2$

Simplify:

$x^2+(y-5)^2=9$

At the point where the two equations intersect, its <em>x-</em>coordinate and <em>y-</em>coordinate will be the same. Therefore, we can substitute the equation of the line into the equation of the circle and solve for <em>x</em>. So:

$x^2+((0.5x+5)-5)^2=9$

Simplify:

$x^2+(0.5x)^2=9$

Square:

$x^2+0.25x^2=9$

Combine like terms:

$\displaystyle 1.25x^2=\frac{5}{4}x^2=9$

Solve for <em>x: </em>

<em /> $\displaystyle \begin{aligned} x^2&=\frac{36}{5} \\ x&=\pm\sqrt{\frac{36}{5}} \\ x&\Rightarrow \frac{6}{\sqrt{5}}=\frac{6\sqrt{5}}{5}\approx2.68\end{aligned}$ <em />

Note that since we are looking for the point of intersection in QI, <em>x</em> should be positive. So, we can ignore the negative answer.

To find the <em>y-</em>coordinate, substitute the <em>x-</em>value back into either equation. Using the linear equation:

$\displaystyle y=0.5\left(\frac{6\sqrt{5}}{5}\right)+5=\frac{3\sqrt{5}}{5}+5\approx 6.34$

So, the point of intersection in QI is:

$\displaystyle \left(\frac{6\sqrt{5}}{5}, \frac{3\sqrt{5}}{5}+5\right)\approx \left(2.68, 6.34)$

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