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3 years ago

In ABC, angle C is a right angle. Which statement must be true

Mathematics

1 answer:

KatRina [158]3 years ago

4 0

Answer:

Step-by-step explanation:

a because the the sin is opposite divided by hypotenuse and the opposite of a is 12 and the hypotenuse is 13 so sin a would be 12/13

this is equals to the cosine of b because cos is adjacent/hypotenuse and the adjacent of b is 12 and the hypotenuse is 13 so cos b would be 12/13

hope this helps :)

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7th grade math plss helpp

ddd [48]

Answer:

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3 years ago

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I need to calculate X 2^x=200 Solve for x

antoniya [11.8K]

Remember
$log(a^z)=zlog(a)$
so

notice that 200=2*2*2*5*5, so x cannot be a whole number

take the log base 10 of both sides

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divide both sides by $log_{10}(2)$
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if we used $log_{2}$ instead of $log_{10}$ we would get
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4 years ago

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Hitman42 [59]

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3 years ago

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Shelia's measured glucose level one hour after a sugary drink varies according to the normal distribution with μ = 117 mg/dl and

TiliK225 [7]

Answer:

The level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L is L = 127.1 mg/dl.

Step-by-step explanation:

To solve this problem, 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.

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 $\frac{\sigma}{\sqrt{n}}$