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3 years ago

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Shawndra said that it is not possible to draw a trapezoid that is a rectangle, but it is possible to draw a square that is a rec

tangle. Marcella said that it is possible. Who is correct? Explain your answer using the properties of quadrilaterals.

2 answers:

ICE Princess25 [194]3 years ago

8 0

Shawndra is correct

She made two statements, and both are true:

1. It is not possible to draw a trapezoid that is a rectangle.

This is true because a trapezoid<span> is a quadrilateral that has exactly one pair of parallel sides, whereas a rectangle is a parallelogram (i.e. it has two pairs of parallel sides)</span>

2. It is possible to draw a square that is a rectangle.

This is true because a rectangle refers to any parallelogram with right angles. A square is also a parallelogram (has two pairs of opposite sides) with right angles. In fact, all squares are rectangles; only that they are a special kind of rectangle, where all the sides are equal in length.

Mnenie [13.5K]3 years ago

5 0

Answer:

Shawndra is correct

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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3 years ago

Need help with b and c

xz_007 [3.2K]

Answer:

Step-by-step explanation:

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3 years ago

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Pls solve this qn..............ASAP<br> PLS DON'T SPAMMMMM!!

Anni [7]

Probability ( a flowering plant or fruit plant) is $\frac{13}{30}$ and Probability (neem plant or peepal plant) is $\frac{29}{60}$ .

Step-by-step explanation:

Given,

Number of neem plant = 125

Number of peepal plant = 165

Number of creppers plant = 50

Number of fruit plant = 150

Number of flowering plant = 110

To find the probability ( a flowering plant or fruit plant)

and probability ( neem or peepal)

Formula

P( A or B) = P(A) + P(B)

Probability ( a flowering plant or fruit plant) = $\frac{110}{600} + \frac{150}{600}$

= $\frac{260}{600}$ = $\frac{13}{30}$

And,

Probability (neem plant or peepal plant) = $\frac{125}{600} +\frac{165}{600}$

= $\frac{290}{600}$ = $\frac{29}{60}$

4 0

3 years ago

Seventy-two percent of the light aircraft that disappear while in flight in a certain country are subsequently discovered. Of th

Anton [14]

Answer:

a) 0.105 = 10.5% probability that it will not be discovered if it has an emergency locator.

b) 0.522 = 52.2% probability that it will be discovered if it does not have an emergency locator.

c) 0.064 = 6.4% probability that 7 of them are discovered.

Step-by-step explanation:

For itens a and b, we use conditional probability.

For item c, we use the binomial distribution along with the conditional probability.

Conditional Probability

We use the conditional probability formula to solve this question. It is

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

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.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

a) If it has an emergency locator, what is the probability that it will not be discovered?

Event A: Has an emergency locator

Event B: Not located.

Probability of having an emergency locator:

66% of 72%(Are discovered).

20% of 100 - 72 = 28%(not discovered). So

$P(A) = 0.66*0.72 + 0.2*0.28 = 0.5312$

Probability of having an emergency locator and not being discovered:

20% of 28%. So

$P(A cap B) = 0.2*0.28 = 0.056$

Probability:

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.056}{0.5312} = 0.105$

0.105 = 10.5% probability that it will not be discovered if it has an emergency locator.

b) If it does not have an emergency locator, what is the probability that it will be discovered?

Probability of not having an emergency locator:

0.5312 of having. So

$P(A) = 1 - 0.5312 = 0.4688$

Probability of not having an emergency locator and being discovered:

34% of 72%. So

$P(A \cap B) = 0.34*0.72 = 0.2448$

Probability:

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.2448}{0.4688} = 0.522$

0.522 = 52.2% probability that it will be discovered if it does not have an emergency locator.

c) If we consider 10 light aircraft that disappeared in flight with an emergency recorder, what is the probability that 7 of them are discovered?

p is the probability of being discovered with the emergency recorder:

0.5312 probability of having the emergency recorder.

Probability of having the emergency recorder and being located:

66% of 72%. So

$P(A \cap B) = 0.66*0.72 = 0.4752$

Probability of being discovered, given that it has the emergency recorder:

$p = P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.4752}{0.5312} = 0.8946$

This question asks for P(X = 7) when $n = 10$ . So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 7) = C_{10,7}.(0.8946)^{7}.(0.1054)^{3} = 0.064$

0.064 = 6.4% probability that 7 of them are discovered.

8 0

3 years ago

Ethan correctly 80% of the total questions on his history Test. He correctly answers 32 question how many questions are on Ethan

natka813 [3]

40 questions because 80% of 40 is 32

3 0

3 years ago

Read 2 more answers