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kondor19780726 [428]
3 years ago
5

ABCD=STQR. What is CD?

Mathematics
2 answers:
ololo11 [35]3 years ago
8 0

Answer:

CD = QR because abcd= star then last 2 are answer

weeeeeb [17]3 years ago
6 0

Answer:

that's the answer

Step-by-step explanation:

cd =qr

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I also got stuck on this one.
Rama09 [41]

Answer:

8 questions.

Step-by-step explanation:

100 - 2x = 84

-2x + 100 = 84

-x + 50 = 42

x - 50 = -42

x = 8

Trevor answered 8 questions incorrectly.

Hope this helps!

6 0
3 years ago
Read 2 more answers
Here is a growing pattern of squares:
Flura [38]

Answer:

squares in Step n. f (n) = 8 + 3(n – 1} for n > 1 /(1) = 8, /{n2) = 3+f (n – 1) for n > 2 01)= 8, 7 (n) = 8= ƒ(n=1) forn> 2 Df1)= 3 -8 (n- 1) forn > 1 Of (n) - 37 + 5 for n > 1 32+5 for n>1 CS (n) 3+ an forn 1

Step-by-step explanation:

3 0
2 years ago
What is the area of the parallelogeam
nlexa [21]
I believe it is 2 feet because area is base multiplied by height. So, 3 x 2/3 is 2.
5 0
4 years ago
56· 328+162· 44+166· 44−x =0<br> find x<br> the dot is to mutiply
Slav-nsk [51]

Answer:

x = 32800

Step-by-step explanation:

56 × 328 + 162 × 44 + 166 × 44 - x = 0

18368 + 7128 + 7304 − x = 0

32800 - x = 0

-x = -32800

x = 32800

8 0
2 years ago
Read 2 more answers
Suppose that a box contains one fair coin (Heads and Tails are equally likely) and one coin with Heads on each side. Suppose fur
stealth61 [152]

Using conditional probability, it is found that there is a 0.1 = 10% probability that the chosen coin was the fair coin.

Conditional Probability

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

  • P(B|A) is the probability of event B happening, given that A happened.
  • P(A \cap B) is the probability of both A and B happening.
  • P(A) is the probability of A happening.

In this problem:

  • Event A: Three heads.
  • Event B: Fair coin.

The probability associated with 3 heads are:

  • 0.5^3 = 0.125 out of 0.5(fair coin).
  • 1 out of 0.5(biased).

Hence:

P(A) = 0.125 + 0.5 = 0.625

The probability of 3 heads and the fair coin is:

P(A \cap B) = 0.5(0.125) = 0.0625

Then, the conditional probability is:

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.0625}{0.625} = 0.1

0.1 = 10% probability that the chosen coin was the fair coin.

A similar problem is given at brainly.com/question/14398287

4 0
3 years ago
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