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2 years ago

True or false if two angles are congruent then they can be moved so that they line up perfectly

Mathematics

1 answer:

tatiyna2 years ago

3 0

If it is what i am thinking, your answer will be TRUE

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The mean radius of earth is 6,371.0 kilometers and the mean radius of Earth's Moon is 1,737.5 kilometers.What is the approximate

Dmitrij [34]

D≈2π(6371-1737.5)

d≈9267π

d≈29113.1km (to nearest tenth of a kilometer)

4 0

3 years ago

A rectangle has a perimeter of 26 units. The length of the rectangle is 2X+3 and the width is 3X. What is the width of the recta

densk [106]

Answer:

X = 2

Step-by-step explanation:

Divided 26 (total perimeter) by 4 = maximum side length possible = 26÷4 =.6.5

I used the trial and error technique

Guessed it was 2 working out to see if i was right or wrong:

Length: 2(2) + 3 = 4+3 = 7

Width: 3(2) = 6

7+7+6+6 = 26

Therefore X = 2

3 0

3 years ago

How do you solve 5 3/7 ÷ 5/11

Artyom0805 [142]

Answer:

11 33/35

Step-by-step explanation:

5 3/7 = 5 × 7/7 + 3/7 = 35 + 37 = 38/7

38/7 ÷ 5/11 = 38/7 × 11/5 = 38 × 11/7 × 5 = 418/35

418/35 = 11 33/35

Hope this helps!

6 0

3 years ago

20. The surface areas of two similar solids are 216 m² and 1014 m². The volume of the larger one is 2197 m³. What is the volume

Andrei [34K]

Answer:

216 m³

Step-by-step explanation:

The ratio of linear dimensions is the square root of the ratio of area dimensions.

s = √(216/1014) = √(36/169) = 6/13

Then the ratio of volume dimensions is the cube of that. The smaller volume is ...

v = (6/13)³·2197 m³ = 216/2197·2197 m³ = 216 m³

The volume of the smaller solid is 216 m³.

8 0

3 years ago

A student S is suspected of cheating on exam, due to evidence E of cheating being present. Suppose that in the case of a cheatin

Kay [80]

Answer:

Step-by-step explanation:

Suppose we think of an alphabet X to be the Event of the evidence.

Also, if Y be the Event of cheating; &

Y' be the Event of not involved in cheating

From the given information:

$P(\dfrac{X}{Y}) = 60\% = 0.6$

$P(\dfrac{X}{Y'}) = 0.01\% = 0.0001$

$P(Y) = 0.01$

Thus, $P(Y') \ will\ be = 1 - P(Y)$

P(Y') = 1 - 0.01

P(Y') = 0.99

The probability of cheating & the evidence is present is = P(YX)

$P(YX) = P(\dfrac{X}{Y}) \ P(Y)$

$P(YX) =0.6 \times 0.01$

$P(YX) =0.006$

The probabilities of not involved in cheating & the evidence are present is:

$P(Y'X) = P(Y') \times P(\dfrac{X}{Y'})$

$P(Y'X) = 0.99 \times 0.0001 \\ \\ P(Y'X) = 0.000099$

(b)

The required probability that the evidence is present is:

P(YX or Y'X) = 0.006 + 0.000099

P(YX or Y'X) = 0.006099

(c)

The required probability that (S) cheat provided the evidence being present is:

Using Bayes Theorem

$P(\dfrac{Y}{X}) = \dfrac{P(YX)}{P(Y)}$

$P(\dfrac{Y}{X}) = \dfrac{P(0.006)}{P(0.006099)}$

$P(\dfrac{Y}{X}) = 0.9838$

5 0

2 years ago