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3 years ago

The mean temperature for the first 4 days in January was -12°C.

Mathematics

1 answer:

tamaranim1 [39]3 years ago

3 0

Answer:

Step-by-step explanation:

Given

The mean temperature for the first 4 days = -12°C
The mean temperature for the first 5 days = -13°C

To find

The temperature on the 5th day

Solution

Total of 4 days = 4*(-12) = -48°C
Total of 5 days = 5*(-13) = -65°C

The difference is the temperature on the 5th day:

-65 - (-48) = -17°C

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C =tid circumference = 3.14 x diameter

chubhunter [2.5K]

Answer:

75.36.

Step-by-step explanation:

C= Pi D (Pi times diameter) so you do 3.14x 24 which then you get your answer of 75.36.

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3 years ago

What is 3/5 × 4/9 as a fraction. Plz help :(

lozanna [386]

3/5 * 4/9 = 4/15. Hope this helps

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3 years ago

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In a coastal town, the humidity, measured in grams of water per kilogram of air, increases by 43% for every 1 degree celsius inc

larisa86 [58]

Answer:

In order to increase humidity by 49.17 grams per kilogram, the increase in temperature will be 18 degree Celsius approx.

Step-by-step explanation:

Initial humidity = 5.75 g/kg

Final humidity= 49.17 g/kg

Percentage increase in humidity= (49.17-5.75)/5.75 *100= 755.13

Increase in temperature= 755.13/43 = 17.56

Increase in temperature = 18 degree Celsius approx.

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3 years ago

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For each statement, write what would be assumed and what would be proven in a proof by contrapositive of the statement. Then wri

Anna007 [38]

Answer:

Given Statement - If x and y are a pair of consecutive integers, then x and y have opposite parity.

Proof by Contrapositive:

Assumed statement: Suppose that integers x and y do not have opposite parity.

Proven Statement: x and y are not a pair of consecutive integers.

Proof -

x = 2u₁ , y = 2u₂

Then

(x, x+1) = (2u₁ , 2u₁ + 1) = (Even, odd)

If y = 2u₁ + 1

Not possible

⇒x and y are not a pair of consecutive integers.

Hence proved.

Proof by Contradiction:

Assumed statement: Suppose x and y are not a pair of consecutive integers.

Proven Statement: Suppose x and y do not have opposite parity.

Proof -

If x and y are not a pair of consecutive integers.

⇒ either x and y are odd or even

If x and y are odd

⇒x and y have same parity

Contradiction

If x and y are even

⇒x and y have same parity

Contradiction

(b)

Proof by Contrapositive:

Assumed statement: Let n be an integer such that n is not odd (i.e. n is an even integer)

Proven Statement: n² is not odd (i.e n² is even)

Proof -

Let n is even

⇒n = 2m

⇒n² = (2m)² = 4m²

⇒n² is even

Hence proved.

Proof by Contradiction:

Assumed statement: Let n be an integer such that n² be odd.

Proven Statement: suppose that n is not odd (i.e n is even)

Proof -

Let n² is odd

⇒n² is even

⇒n² = 2m

⇒2 | n²

⇒2 | n

⇒n = 2x

⇒ n is even

Contradiction

8 0

3 years ago

You know that in a specific population of rainbow trout 15% of the individuals carry intestinal parasites. Assume you obtain a r

kicyunya [14]

Answer:

a) 0.2316 = 23.16% probability that 0 carry intestinal parasites.

b) 0.4005 = 40.05% probability that at least two individuals carry intestinal parasites.

Step-by-step explanation:

For each trout, there are only two possible outcomes. Either they carry intestinal parasites, or they do not. Trouts are independent. This means that we use the binomial probability distribution to solve this question.

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.

You know that in a specific population of rainbow trout 15% of the individuals carry intestinal parasites.

This means that $p = 0.15$

Assume you obtain a random sample of 9 individuals from this population:

This means that $n = 9$

a. Calculate the probability that __ (last digit of your ID number) carry intestinal parasites.

Last digit is 0, so:

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

$P(X = 0) = C_{9,0}.(0.15)^{0}.(0.85)^{9} = 0.2316$

0.2316 = 23.16% probability that 0 carry intestinal parasites.

b. Calculate the probability that at least two individuals carry intestinal parasites.

This is

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1)$

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

$P(X = 0) = C_{9,0}.(0.15)^{0}.(0.85)^{9} = 0.2316$

$P(X = 1) = C_{9,1}.(0.15)^{1}.(0.85)^{8} = 0.3679$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.2316 + 0.3679 = 0.5995$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.5995 = 0.4005$

0.4005 = 40.05% probability that at least two individuals carry intestinal parasites.

5 0

3 years ago