Find the consecutive positive odd integers whose product is 99

Mathematics

1 answer:

dezoksy [38]3 years ago

4 0

Answer:

9 and 11

Step-by-step explanation:

So let's say that the first integer is x.

That means that the second integer is x+2, since it us the next odd number.

This problem can be easily solved with our mind the answer is 9 and 11, but I'll show you the steps to slove this.

We make an equation using these two numbers:

$(x)*(x+2)=99\\$

Now all we gotta do is solve for x:

$x^2 +2x = 99\\x^2+2x-99=0$

Now we use either splitting the middle term or quadratic formula:

$x^2+11x-9x+99=0\\x(x+11)-9(x+11)=0\\(x+11)(x-9)=0$

Now we split each terma and solve for x:

$(x-9)=0\\x=9\\(x+11)=0\\x=-11$

Now the question states <em>positive </em>so we can rule out x = -11.

Now we have the first integer x = 9,

the second integer is x+2 = 9+2 =11

So the two consecutive positive odd integers are 9 and 11

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andrey2020 [161]

Answer:

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Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

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The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

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