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3 years ago

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There is a traffic light at the intersection of Main Street and Durham Avenue. The traffic light on Main Street follows a cycle.

It is green for 30 seconds, yellow for 5 seconds, and red for 25 seconds. As you travel along Main and approach the intersection, what is the probability that the first color you see is green?

1 answer:

givi [52]3 years ago

6 0

Probability is defined as the number of desired outcomes over all possible outcomes. Lets say each second is one possible outcome. This means there is 30 different desired outcomes out of 60 possible outcomes. The probability therefore is 30/60, simplified to 1/2.

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Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

melamori03 [73]

Answer:

$6+2\sqrt{21}\:\mathrm{cm^2}\approx 15.17\:\mathrm{cm^2}$

Step-by-step explanation:

The quadrilateral ABCD consists of two triangles. By adding the area of the two triangles, we get the area of the entire quadrilateral.

Vertices A, B, and C form a right triangle with legs $AB=3$ , $BC=4$ , and $AC=5$ . The two legs, 3 and 4, represent the triangle's height and base, respectively.

The area of a triangle with base $b$ and height $h$ is given by $A=\frac{1}{2}bh$ . Therefore, the area of this right triangle is:

$A=\frac{1}{2}\cdot 3\cdot 4=\frac{1}{2}\cdot 12=6\:\mathrm{cm^2}$

The other triangle is a bit trickier. Triangle $\triangle ADC$ is an isosceles triangles with sides 5, 5, and 4. To find its area, we can use Heron's Formula, given by:

$A=\sqrt{s(s-a)(s-b)(s-c)}$ , where $a$ , $b$ , and $c$ are three sides of the triangle and $s$ is the semi-perimeter ( $s=\frac{a+b+c}{2}$ ).

The semi-perimeter, $s$ , is:

$s=\frac{5+5+4}{2}=\frac{14}{2}=7$

Therefore, the area of the isosceles triangle is:

$A=\sqrt{7(7-5)(7-5)(7-4)},\\A=\sqrt{7\cdot 2\cdot 2\cdot 3},\\A=\sqrt{84}, \\A=2\sqrt{21}\:\mathrm{cm^2}$

Thus, the area of the quadrilateral is:

$6\:\mathrm{cm^2}+2\sqrt{21}\:\mathrm{cm^2}=\boxed{6+2\sqrt{21}\:\mathrm{cm^2}}$

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3 years ago

A fair dice is rolled

iren2701 [21]

Answer:

1/6

Step-by-step explanation:

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3 years ago

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azamat

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3 years ago

Read 2 more answers

If 2k, 5k-1 and 6k+2 are the first 3 terms of an arithmetic sequence, find k and the 8th term.

Simora [160]

Consecutive terms in an arithmetic sequence differ by a constant $d$ . So

$5k-1=2k+d$
$6k+2=5k-1+d$
$\implies\begin{cases}3k-d=1\\k-d=-3\end{cases}\implies k=2,d=5$

Denote the $n$ -th term in the sequence by $a_n$ . Now that $a_1=2$ , we have

$a_n=a_{n-1}+5=a_{n-2}+2\cdot5=\cdots=a_1+(n-1)\cdot5$

which means

$a_8=2+(8-1)\cdot5=37$

4 0

3 years ago

A tire manufacturer warranties its tires to last at least 20,000 miles orâ "you get a new set ofâ tires." In itsâ experience, a

Y_Kistochka [10]

Answer:

Probability that a set of tires wears out before 20,000 miles is 0.1151.

Step-by-step explanation:

We are given that a tire manufacturer warranties its tires to last at least 20,000 miles or "you get a new set of tires." In its past experience, a set of these tires last on average 26,000 miles with S.D. 5,000 miles. Assume that the wear is normally distributed.

<em>Let X = wearing of tires</em>

So, X ~ N( $\mu=26,000,\sigma^{2}=5,000^{2}$ )

Now, the z score probability distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = average lasting of tires = 26,000 miles

$\sigma$ = standard deviation = 5,000 miles

So, probability that a set of tires wears out before 20,000 miles is given by = P(X < 20,000 miles)

P(X < 20,000) = P( $\frac{X-\mu}{\sigma}$ < $\frac{20,000-26,000}{5,000}$ ) = P(Z < -1.2) = 1 - P(Z $\leq$ 1.2)

= 1 - 0.88493 = 0.1151

Therefore, probability that a set of tires wears out before 20,000 miles is 0.1151.

4 0

3 years ago