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3 years ago

I need help solving for X

Mathematics

2 answers:

8_murik_8 [283]3 years ago

6 0

Answer:

x = 63

Step-by-step explanation:

This is a right angle, and we know that there are 90 total degrees in a right angle. Because of this information, we know that the degrees of both angles, added together, should come out to 90. This reasoning looks like this:

$25+(x+2)=90$

Now, it's much easier to solve for x. I'll start by removing the parentheses and combining like terms.

$25+x+2=90\\25+2+x=90\\27+x=90$

Now, let's remove 27 both sides to isolate x and figure out what it's equal to.

$27+x=90\\x = 63$

So, the measure of x is 63 degrees. We can make sure this is correct by using our equation:

Is $25+(63+2)=90$ ? Yes, because 63 + 2 is equal to 65, and 65 plus 25 is equal to 90. So our answer is correct.

If you need any more help, let me know! :)

finlep [7]3 years ago

4 0

Answer:

x = 63°

Step-by-step explanation:

(x + 2) + 25 = 90

x + 2 = 90 - 25

x + 2 = 65

x = 65 - 2

x = 63°

(x + 2) + 25 = 90

(63 + 2) + 25 = 90

65 + 25 = 90

90 = 90

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4 years ago

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Answer:

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4 years ago

Sylvia buys a case of water that contains 24 bottles and sells x number of bottles to her co-workers. Sam buys a case of water t

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Answer:

Step-by-step explanation:

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3 years ago

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Step-by-step explanation:

To solve this problem, it is important to know the normal probability distribution and the central limit theorem.

Normal probability distribution

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In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$