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A basketball diamond is actually a square with sides of 90 feet. If a runner tries to steal second base, how far must the catche

Tpy6a [65]

Answer:

127.28 feet

Step-by-step explanation:

Given

Shape of basketball diamond = Square

Length of a side = 90 feet

If the length of a side is 90 feet then the length of other sides is 90 feet.

By finding how far the catcher, at home plate, throw to get the runner "out", we are basically finding the distance from 2nd base to home

This means we're calculating the diagonal of the field..

(Since the field is a square, it has a right angle at 1st base)

Let the diagonal be representing by x.

From Pythagoras,

x² = 90² + 90²

x² = 8100 + 8100

x² = 16200

x = √16200

x = 127.2792206135785

x = 127.28 feet ------ Approximated

4 0

3 years ago

4. In a sample of 1000 people, 130 can wiggle their ears. Two unrelated people are

Elina [12.6K]

Using the hypergeometric distribution, it is found that there is a 0.7568 = 75.68% probability that neither can wiggle his or her ears.

The people are chosen from the sample without replacement, which is why the <u>hypergeometric distribution</u> is used to solve this question.

Hypergeometric distribution:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.
N is the size of the population.
n is the size of the sample.
k is the total number of desired outcomes.

In this problem:

1000 people means that $N = 1000$
130 can wiggle their ears, thus $k = 130$
Two are selected, thus $n = 2$ .

The probability that neither can wiggle his or her ears is P(X = 0), thus:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}$

$P(X = x) = h(0,1000,2,130) = \frac{C_{130,0}C_{870,2}}{C_{1000,2}} = 0.7568$

0.7568 = 75.68% probability that neither can wiggle his or her ears.

A similar problem is given at brainly.com/question/24826394

5 0

3 years ago

Use the box plots comparing the number of boy dogs and number of girl dogs attending doggy daycare each day for a month to answe

satela [25.4K]

The boxplots 1 and 2 represent the distribution of data using a five number summary for the male and the female attendance

The first box plot represents the male's boxplot.

So, we have:

Upper quartile (Q3) = 14
Lower quartile (Q1) = 2

The IQR is the difference between the above values

IQR = Q3 - Q1

So, we have:

IQR = 14 - 2 = 12

So: the IQR for the males' data is 12

<h3>(b) The difference in the median values</h3>

From the first and second boxplots, we have:

Median male = 10
Median female = 18

The difference (d) between these values is

d = 18 - 10

d = 8

So: the difference between the median values of each data set is 8

<h3>(c) The distribution of data</h3>

This is dependent on whether the dataset has an outlier or not.

From the figure, we have:

Male's data: presence of an outlier - use median
Female's data: absence of an outlier - use mean

So: the male dataset is measured by median, while the mean is a better measure of center for the female's data

<h3>(d) Reason for outliers</h3>

A possible reason for outliers is sampling problems