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2 years ago

What is the diameter of circle T?

Mathematics

1 answer:

Ratling [72]2 years ago

5 0

Answer:

$Diameter = 16$

Step-by-step explanation:

Given

$\theta = 72$

$Sector = \frac{64}{5}\pi$

Required

Calculate the diameter

The area of a sector is:

$Area = \frac{\theta}{360} * \pi r^2$

Substitute values for Area and $\theta$

$\frac{64}{5}\pi = \frac{72}{360} * \pi r^2$

Divide both sides by $\pi$

$\frac{64}{5} = \frac{72}{360} * r^2$

Make $r^2$ the subject

$\frac{64*360}{5*72} = r^2$

$\frac{64*360}{360} = r^2$

$64 = r^2$

Take positive square roots of both sides

$\sqrt{64} = r$

$8 = r$

$r = 8$

The diameter is then calculated as:

$Diameter = 2r$

$Diameter = 2*8$

$Diameter = 16$

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20 PTS What is the sixth term of the sequence shown in the table? Term Value 1 –46,656 2 7,776 3 –1,296 4 216 5 6 {SEE IMAGE}

olganol [36]

Answer:

a₆ = 6

Step-by-step explanation:

There is a common ratio r between consecutive terms in the sequence, that is

r = $\frac{216}{-1296}$ = $\frac{-1296}{7776}$ = $\frac{7776}{-46656}$ = - $\frac{1}{6}$

This indicates the sequence is geometric.

To obtain any term in the sequence, multiply the previous term by r, thus

a₅ = 216 × - $\frac{1}{6}$ = - 36

a₆ = - 36 × - $\frac{1}{6}$ = 6

8 0

3 years ago

P(X< ) 1-P(X> ) A softball pitcher has a 0.626 probability of throwing a strike for each curve ball pitch. If the softball

Mashutka [201]

Answer:

19.49% probability that no more than 16 of them are strikes

Step-by-step explanation:

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 $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .