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melisa1 [442]
3 years ago
14

Geometry help greatly needed. please help if possible.

Mathematics
1 answer:
elixir [45]3 years ago
3 0

Answer:

\huge\boxed{\sf  x = 90\°}

Step-by-step explanation:

If AB is diameter, the angle opposite to it, when a triangle is inscribed in that semi circle, will be 90 degrees.

<u>Given that:</u>

AB is the diameter.

According to what is given, we come to know that <ACB = x = 90°.

\rule[225]{225}{2}

Hope this helped!

<h3>~AH1807</h3>
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What is the slope of the line?
myrzilka [38]

Answer:

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7 0
3 years ago
20% of US High School teens vape. A local High School has implemented campaigns to reduce vaping among students and believes tha
zaharov [31]

Answer:

10.93% probability of observing 51 or fewer vapers in a random sample of 300

Step-by-step explanation:

I am going to use the normal approximation to the binomial to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be 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.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

n = 300, p = 0.2

So

\mu = E(X) = np = 300*0.2 = 60

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{300*0.2*0.8} = 6.9282

What is the approximate probability of observing 51 or fewer vapers in a random sample of 300?

Using continuity corrections, this is P(X \leq 51 + 0.5) = P(X \leq 51.5), which is the pvalue of Z when X = 51.5 So

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

Z = \frac{51.5 - 60}{6.9282}

Z = -1.23

Z = -1.23 has a pvalue of 0.1093.

10.93% probability of observing 51 or fewer vapers in a random sample of 300

4 0
3 years ago
The table shows the outputs, y, for different inputs, x:
ASHA 777 [7]
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Yes, the data represent a function because there is at least one x-value for every y-value.

Part B:
When x=6 in the input-output table, y=14. When x=6 in the relation f(x)=7x-15, f(x)=7(6)-15=27. <span>The equation has a greater value when x=6.

Part C:</span>
Set f(x) equal to 6 in the equation:
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Solve for x:
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