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A 1,260-foot tree has grown at a constant rate each year. In the equation below, t is the age of the tree 28t = 1,260 in years.

prohojiy [21]

Answer:

45 years

Step-by-step explanation:

28t = 1,260

Divide both sides by 28

t = 45

4 0

3 years ago

From a point on a circle, two perpendicular chords are drawn. One is 6 cm from the center and the other one is 3 cm from the cen

Ira Lisetskai [31]

The chord 3cm away is 12cm and chord 6cm away is 6cm.
The two perpendicular chords will create a rectangle with the center. each side will be half of the parallel sides length.

4 0

3 years ago

Read 2 more answers

approximate the integral 0 to pi/2 xsin(x)dx by computing the Left and Right sums, using the parition (0, pi/6, pi/4, pi/3, pi/2

Cerrena [4.2K]

3.14 times all will give you the answer

8 0

3 years ago

Suppose data made available through a health system tracker showed health expenditures were $10,348 per person in the United Sta

nignag [31]

Answer:

a) 30.08% probability the sample mean will be within $100 of the population mean.

b) 0% probability the sample mean will be greater than $12,600

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the Central Limit Theorem.

Normal probability distribution

When the distribution is normal, we use 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 normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .