Help in the triangle QUESTION plz :'(

The period, T, of a pendulum of length l is given by T = f(l) = 2π√l/g, where g is a constant. Find f^-1.

snow_lady [41]

Answer:

When effective length (L) is constant, the time period (T) of oscillation of a simple pendulum is inversely proportional to the square root of the acceleration due to gravity (g) at a place of observation. So, T∝√1g when L is constant.

Step-by-step explanation:

5 0

4 years ago

The water works commission needs to know the mean household usage of water by the residents of a small town in gallons per day.

ozzi

Answer:

The 80% confidence interval for the mean usage of water is between 18.4 and 18.6 gallons per day.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.8}{2} = 0.1$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.1 = 0.90$ , so $z = 1.28$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.28*\frac{2.3}{\sqrt{717}} = 0.1$

The lower end of the interval is the sample mean subtracted by M. So it is 18.5 - 0.1 = 18.4 gallons per day.

The upper end of the interval is the sample mean added to M. So it is 18.5 + 0.1 = 18.6 gallons per day.

The 80% confidence interval for the mean usage of water is between 18.4 and 18.6 gallons per day.

6 0

3 years ago

The cost of 5 gallons of ice cream has a standard deviation of 8 dollars with a mean of 29 dollars during the summer. What is th

Feliz [49]

Answer:

97.74% probability that the sample mean would differ from the true mean by less than 1.9 dollars if a sample of 92 5-gallon pails is randomly selected

Step-by-step explanation:

To solve this question, 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 $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 29, \sigma = 8, n = 92, s = \frac{8}{\sqrt{92}} = 0.8341$

What is the probability that the sample mean would differ from the true mean by less than 1.9 dollars if a sample of 92 5-gallon pails is randomly selected?

This is the pvalue of Z when X = 29 + 1.9 = 30.9 subtracted by the pvalue of Z when X = 29 - 1.9 = 27.1. So

X = 30.9

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

By the Central Limit Theorem

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

$Z = \frac{30.9 - 29}{0.8341}$

$Z = 2.28$

$Z = 2.28$ has a pvalue of 0.9887

X = 27.1

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

$Z = \frac{27.1 - 29}{0.8341}$

$Z = -2.28$

$Z = -2.28$ has a pvalue of 0.0113

0.9887 - 0.0113 = 0.9774

97.74% probability that the sample mean would differ from the true mean by less than 1.9 dollars if a sample of 92 5-gallon pails is randomly selected

7 0

3 years ago

Write the number with the same value as a 28 tens

exis [7]

28 tens = 2,800 so 2,800 ones

7 0

3 years ago

Read 2 more answers

Which list shows the weight of the snacks in pounds in order from lightest to heaviest?

kari74 [83]

Answer:

where is the list?????