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aliina [53]
2 years ago
13

4. Rhombus QRST has diagonals intersecting at W. Point U is located on side QR and point V on diagonal RT

Mathematics
1 answer:
ad-work [718]2 years ago
8 0

Answer:

Given:

In Rhombus QRST, diagonals QS and RT intersect at W and U∈QR and point V∈RT  such that UV⊥QR. (shown in below diagram)

To prove: QW•UR =WT•UV

Proof:

In a rhombus diagonals bisect perpendicularly,

Thus, in QRST

QW≅WS, WR ≅ WT and m∠QWR=m∠QWT=m∠RWS=m∠TWS=90°.

In triangles QWR and UVR,

m\angle QWR=m\angle VUR              (Right angles)

m\angle WRQ=m\angle VRU              (Common angles)

By AA similarity postulate,

\triangle QWR\sim \triangle VUR

The corresponding sides in similar triangles are in same proportion,

\implies \frac{QW}{VU}=\frac{WR}{UR}

QW\times UR=WR\times VU

QW\times UR=WT\times UV                 (∵ WR ≅ WT )

Hence, proved.

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Answer:

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A: 5 + 8 = 13

B: 6 + 7 = 13

C: 7 + 2 = 9

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D: 8 + 8 = 16.

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What is the answer to 2X +1=-15
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<span>2X +1=-15
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Harrison Water Sports has two retail outlets: Seattle and Portland. The Seattle store does 60 percent of the total sales in a ye
Anna11 [10]

Answer: P = 0.75

Step-by-step explanation:

Hi!

The sample space of this problems is the set of all the possible sales. It is divided in the disjoint sets:

S_s = {\text{sales made in Seattle }}\\S_p={\text {sales made in Portland}}

We have also the set of sales of boat accesories S_b, the colored one in the image.

We are given the data:

P(S_s) = 0.6\\P(S_b | S_s) = \frac{P(S_b\bigcap S_s)}{P(S_s)}=0.4\\P(S_b|S_p) =\frac{P(S_b\bigcap S_p)}{P(S_p)}=0.2

From these relations you can compute the probabilities of the intersections colored in the image:

pink\;set:\;P(S_b \bigcap S_s) =0.6*0.4=0.24\\blue\;set\;:P(S_b \bigcap S_p)=(1-0.6)*0.2 =0.08

You are asked about the conditional probability:

P(S_s|S_b) = \frac{P(S_s \bigcap S_b)}{P(S_b)}

To calculate this, you need  P(S_b) . In the image you can see that the set S_b is the union of the two disjoint pink and blue sets. Then:

P(S_b)=P((S_b \bigcap S_s)\bigcup(S_b \bigcap S_p)) = 0.24 + 0.08 = 0.32

Finally:

P(S_s|S_b) = \frac{0.24}{0.32}=0.75

4 0
2 years ago
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