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meriva
3 years ago
6

Two rhombuses with the same side lengths are always congruent ?

Mathematics
1 answer:
lyudmila [28]3 years ago
8 0

Answer:

False

Step-by-step explanation:

A rhombus is a quadrilateral with

1. all side length equal

2. opposite side parallel

3. opposite angles are congruent.

______________________________________

If there are two rhombus with same side length ,

conditions so that they must be always congruent.

1. Side length must be equal

2. corresponding angles to equal sides must also be equal.

It is possible that the sides of two rhombus will be same

but their angles can differ and hence the two rhombus may or may not be congruent.

Thus,

Two rhombuses with the same side lengths are always congruent

given statement is false

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What is the volume of this sphere?
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Answer:

523.59 or 523.60

Step-by-step explanation:

Volume for a sphere: 4/3πr^3 (4÷3x π x r ^ 3)

(r is radius, V is volume)

V= 4/3 x π x r^3

The diameter is 10, so the radius is 5...

V= 4/3 x π x 5^3

V= 4/3 x π x 125

CALCULATER: 4 ÷ 3 x π x 125

Answer 523.59  (523.60 rounded)

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3 years ago
In a pack of skittles, 27% are red what percentage are not red?
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Answer:

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Step-by-step explanation:

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3 years ago
Read 2 more answers
The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponenti
melomori [17]

Answer:

a) 0.1496 = 14.96% probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day.

b) Capacity of 252.6 cubic feet per second

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

f(x) = \mu e^{-\mu x}

In which \mu = \frac{1}{m} is the decay parameter.

The probability that x is lower or equal to a is given by:

P(X \leq x) = \int\limits^a_0 {f(x)} \, dx

Which has the following solution:

P(X \leq x) = 1 - e^{-\mu x}

The probability of finding a value higher than x is:

P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}

The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponential distribution with mean 100 cfs (cubic feet per second).

This means that m = 100, \mu = \frac{1}{100} = 0.01

(a) Find the probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day. (Round your answer to four decimal places.)

We have that:

P(X > x) = e^{-\mu x}

This is P(X > 190). So

P(X > 190) = e^{-0.01*190} = 0.1496

0.1496 = 14.96% probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day.

(b) What water-pumping capacity, in cubic feet per second, should the station maintain during early afternoons so that the probability that demand will exceed capacity on a randomly selected day is only 0.08?

This is x for which:

P(X > x) = 0.08

So

e^{-0.01x} = 0.08

\ln{e^{-0.01x}} = \ln{0.08}

-0.01x = \ln{0.08}

x = -\frac{\ln{0.08}}{0.01}

x = 252.6

Capacity of 252.6 cubic feet per second

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3 years ago
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Answer:

Step-by-step explanation:

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V_i=1875\pi

Subtract the inner from the outer:

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If you need your answer in decimal form, then the volume is

9189.16 cm³

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Answer:

See attached

Step-by-step explanation:

The proof is attached

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