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3 years ago

How do you change 2/5 to a precent

Mathematics

1 answer:

nikklg [1K]3 years ago

8 0

Answer:

40%

Step-by-step explanation:

Percent is how much out of 100.

We have 2/5, so find equivalent fractions:

2/5 = 4/10 = 40/100

2/5 = 40%

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Complete Question

The complete question is shown on the first uploaded image

Answer:

$P(U |D ) = 0.198$

$P(O\ n \ B) = 0.188$

$P(O | B) = 0.498$

Step-by-step explanation:

The total number of deaths is mathematically represented as

$T = 16 + 23 + \cdots + 16$

$T =623$

The total number of deaths in 1996 - 2000 is mathematically represented as

$T_a = 16 + 23+ \cdots + 30$

$T_a = 235$

The total number of deaths in 2001 - 2005 is mathematically represented as

$T_b = 17 + 16 + \cdots + 23$

$T_b = 206$

The total number of deaths in 2006 - 2010 is mathematically represented as

$T_c = 15 + 17 + \cdots + 16$

$T_d = 182$

Generally the the probability that it would occur under the tree given that the death was after 2000 is mathematically represented as

$P(U |D ) = \frac{P(A \ n\ U )}{P(A)}$

Here $P(A \ n\ U )$ represents the probability that it was after 2000 and it was under the tree and this is mathematically represented as

$P(A \ n\ U ) = \frac{Z}{ T}$

Here Z is the total number of death under the tree after 2000 and it is mathematically represented as

$Z = 35 + 42$

=> $Z = 77$

=> $P(A \ n\ U ) = \frac{77}{ 623}$

Also

$P(A)$ is the probability of the death occurring after 2000 and this is mathematically represented as

$P(A) = \frac{T_b + T_c}{ T}$

=> $P(A) = \frac{ 206+ 182}{623}$

$P(U |D ) = \frac{\frac{77}{ 623} }{ \frac{ 206+ 182}{623}}$

=> $P(U |D ) = 0.198$

Generally the probability that the death was from camping or being outside and was before 2001 is mathematically represented as

$P(O | B) = \frac{T_z}{ T}$

Here $T_z$ is the total number of death outside / camping before 2001 and the value is 117

$P(O \ n \ B) = \frac{117}{623}$

=> $P(O\ n \ B) = 0.188$

Generally the probability that the death was from camping or being outside given that it was before 2001 is mathematically represented as

$P(O | B) = \frac{ P( O \ n \ B)}{ P(B)}$

Here $P(B)$ is the probability that it was before 2001 , this is mathematically represented as

$P(B ) = \frac{T_a}{T}$

=> $P(B ) = \frac{235}{623}$

$P(O | B) = \frac{ \frac{117}{623}}{ \frac{235}{623}}$

=> $P(O | B) = 0.498$

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2 years ago

A farmer has a truckload of watermelons to take to market. At the beginning of the trip, the watermelons weigh 1000 kg. 99% of t

Mariulka [41]

Answer:

The weight of watermelon at the end of the trip is <u>500 kg.</u>

Step-by-step explanation:

Given:

The weight of watermelons is 1000kg.

At the beginning of the trip watermelon is 99% of the weight:

99% of 1000kg

$\frac{99}{100}\times 1000$

$=0.99\times 1000$

$= 990$ kg

Now, at the end of the trip it is 98% of weight. So, let $x$ kg of water has been evaporated. So, remaining weight of the watermelon is $1000-x$ .

If $x$ kg of water is removed, then the amount of water left is $990-x$ .

As per question, water now weighs 98% of the total weight. So,

98% of ( $1000-x$ ) = $990-x$

$\frac{98}{100}\times (1000-x)=990-x$

$980-0.98x=990-x$

$-0.98x+x=990-980$ (Bringing like terms on one side)

$0.02x=10\\x=\frac{10}{0.02}=500\ kg$

Therefore, the weight of watermelon at the end of the trip is given as:

Weight = $1000 - x$ = 1000 - 500 = 500 kg

Therefore, the watermelon weighs 500 kg at the end of trip.

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Answer:

Step-by-step explanation:

The angle whose sine is 5/12 = 22.62 degrees and in Quadrant II it is

180 - 22.62 degrees

The angle whose tan is (5/12) = 22.62.

So we can write α as 180 - α and β as α.

sin (α+β) = sin ( 180 - α = α) = sin 180

= 0.

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