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leonid [27]
3 years ago
5

What is the probability that a stamp chosen at random is a commemorative or a special delivery stamp?

Mathematics
1 answer:
Ksivusya [100]3 years ago
3 0

Answer: Option 'D' is correct.

Step-by-step explanation:

Since we have given that

P(\text{ getting a commemorative stamp})=0.16

and

P(\text{getting a special delivery stamp})=0.18

So, we need to find,

P(\text{getting either a commemorative or a special delivery stamp})\\=P(\text{getting a commemorative stamp})+P({\text{ getting a special delivery stamp})

(∵they are independent events .)

P(\text{getting either a commemorative or a special delivery stamp})\\=0.16+0.18\\=0.34

So, option 'D' is correct.

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Triangle KML and triangle PRQ are shown in the figure belowK404008 cms4 cms10 cmsOL3 cms6 cmsRM17) Are these triangles similar?1
s2008m [1.1K]

Two triangles are similar if the only difference between them is the size, this means that their internal angles must be the same. If we look at the picture the first triangle has one angle equal to 40 degrees, one equal to 80 degrees and the third one is unkown (x). The second triangle has one angle equal to 40 degrees, one equal to 60 degrees and the third one is unkown (y). The sum of the internal angles of a triangle must be equal to 180 degrees, with this information we can find the values of the missing angles. We have:

\begin{gathered} 40+80+x=180 \\ 120+x\text{ = 180} \\ x\text{ = 180-120=60} \end{gathered}\begin{gathered} 40+60+y=180 \\ 100+y=180 \\ y=180-100=80 \end{gathered}

Therefore the internal angles of the first triangle are (40,80,60) and the angles of the second triangle are (40,80,60) as well, therefore they are similar.

Two triangles are congruent if they have sides with the same length. Which is not the case, because the sides of one triangle is (8, 10, 6) while the other is (4,3 and unkown). Therefore they are not congruent.

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1 year ago
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The answer is b, c, and d
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A triangle has one side that is 5 inches long and the other two sides are both of length x. Write an expression for the perimete
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Let P = perimeter

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Es 18 de junio de 1815, las tropas napoleónicas se encuentran justo adelante, tu eres el ingeniero en balística y el general Wel
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Answer:

89.1° or -1.4°  

Step-by-step explanation:

1. Location:

You are on the Mont-Saint-Jean escarpment, near the Belgian town of Waterloo.

The French troops are about 50 m below you and 1.2 km distant.

2. Finding the firing angle

Data:

R = 1200 m

u = 600 m/s

h = -50 m (the height of the target)

a = 9.8 m/s²

We have two conditions.

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Vertical distance

(2) -50 = 600t sinθ - 4.9t²

Divide each side of (1) by 600cosθ.

(3) \, t =\dfrac{2}{\cos \theta}

Substitute (3) into (2)

-50 = 600t \sin \theta - 4.9t^{2} =  600 \left( \dfrac{2}{\cos \theta} \right ) \sin \theta - 4.9 \left( \dfrac{2}{\cos \theta} \right )^{2}\\\\(4) \, -50 = 1200 \tan \theta - \dfrac{19.6}{\cos^{2} \theta}

Recall that

(5) sec²θ = 1/cos²θ = tan²θ + 1

Substitute (5) into (4)

-50 = 1200 \tan \theta - 19.6 \left(\tan^{2} \theta}+ 1\right )

Set up a quadratic equation

\begin{array}{rcl}-50 & = & 1200 \tan \theta - 19.6\tan^{2} \theta -19.6 \\0 & = & 1200 \tan \theta - 19.6\tan^{2} \theta + 30.4\\0 & =&19.6\tan^{2} \theta - 1200 \tan \theta - 30.4\\0 & =&\tan^{2} \theta - 61.224 \tan \theta - 1.551\\\end{array}

Solve for θ

Use the quadratic formula.

tanθ = 61.249 or -0.025

θ = arctan(61.249) = 89.1° or

θ = arctan(-0.025) = -1.4°

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