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yanalaym [24]
3 years ago
10

Find three consecutive odd integers such that the sum of the first and third equals the sum of the second and 33.​

Mathematics
2 answers:
ratelena [41]3 years ago
7 0
Answer:
31, 33, 35

Explanation:
We can set the variable for the first integer of the three consecutive integers as x.

First integer=x

Because they are all odd, we can then say that the second integer is equal to the first integer plus 2.

Second integer=x+2

Using the same knowledge, we can say that the third integer is equal to the second integer plus 2.

Third integer=x+2+2
Third integer=x+4.

Now, we have our three integers:
First integer=x
Second integer=x+2
Third integer=x+4

We can write out equation out like this:
(x)+(x+4)=(x+2)+33
The sum of the first (x) and third (x+4) integers is equal to the sum of the second (x+2) and 33.

Now, we solve this equation.
(x)+(x+4)=(x+2)+33
Open up the parentheses
x+x+4=x+2+33
Combine like terms
2x+4=x+35
Subtract both sides by 4
2x+4-4=x+35-4
2x=x+31
Subtract both sides by x
2x-x=x+31-x
x=31

Now, we know that our first integer is 31. Because these integers are consecutive and odd, we know that our second integer is 33 and the third is 35.

I hope this helps!
Klio2033 [76]3 years ago
5 0

Answer:

31, 33, 35

Step by step explanation:

Due to a lack of time, I can't explain it right now, apologies.

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Simplify 8/5 * -6 to -48/5

Move the negative sign to the left

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Which of these problem types cannot be solved using the law of sines
timama [110]

Answer: The correct option are B, C and D.

Explanation:

The law of sine states that,

\frac{\sin A}{a} =\frac{\sin B}{b}=\frac{\sin C}{c}

Where A, B, C are interior angles of the triangle and a, b, c are sides opposite sides of these angles respectively as shown in below figure. Only AAS or SSA types problems can be solved by using Law of sine.

Since we need the combination of two sides and one angle or two angles and one side.

In option A, the two consecutive angles are known and a side which makes the second angle with base side is known, therefore the first angle is opposite to the given side, so the law of sine can be used for AAS problems.

Therefore, option A is incorrect.

In option B a side is known and two inclined angle on that line are known. But to use Law of sine we want the line and angle which in not inclined on that line, therefore the ASA problem can not be solved by Law of sine and the option B is correct.

In option C two sides and their inclined angle is known. But to use Law of sine we want the side and angle which in not inclined on that line, therefore the SAS problem can not be solved by Law of sine and the option C is correct.

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3 years ago
On a particular game show, there are 8 covered buckets and 2 of them contain a ball.
kobusy [5.1K]

Answer:

0.2143 = 21.43% probability that a contestant wins the game if he/she gets to select 4 of the buckets.

Step-by-step explanation:

The buckets are chosen without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x sucesses is given by the following formula:

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}

In which:

x is the number of sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

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)!}

In this question:

8 covered buckets, so N = 8.

4 buckets are selected, so n = 4.

2 contain a ball, which means that k = 2.

Find the probability that a contestant wins the game if he/she gets to select 4 of the buckets.

This is P(X = 2). So

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}

P(X = 2) = h(2,8,4,2) = \frac{C_{2,2}*C_{6,2}}{C_{8,2}} = 0.2143

0.2143 = 21.43% probability that a contestant wins the game if he/she gets to select 4 of the buckets.

7 0
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