What is 2 + 1 - 3 = 5 3 6

Three points are colinear. always sometimes never

Novosadov [1.4K]

Three or more points are collinear, if slope of any two pairs of points is same. With three points A, B and C, three pairs of points can be formed, they are: AB, BC and AC. If Slope of AB = slope of BC = slope of AC, then A, B and C are collinear points

6 0

3 years ago

The measure of an angle is 60°. What is the measure of its supplementary angle?

SVEN [57.7K]

Answer:

120

Step-by-step explanation:

supplementary equal up 180

3 0

3 years ago

Suppose that a basketball player can score on a particular shot with probability .3. Use the central limit theorem to find the a

Rom4ik [11]

Answer:

(a) The probability that the number of successes is at most 5 is 0.1379.

(b) The probability that the number of successes is at most 5 is 0.1379.

(c) The probability that the number of successes is at most 5 is 0.1379.

(d) The probability that the number of successes is at most 11 is 0.9357.

→ All the exact probabilities are more than the approximated probability.

Step-by-step explanation:

Let S = a basketball player scores a shot.

The probability that a basketball player scores a shot is, P (S) = p = 0.30.

The number of sample selected is, n = 25.

The random variable $S\sim Bin(25,0.30)$

According to the central limit theorem if the sample taken from an unknown population is large then the sampling distribution of the sample proportion ( $\hat p$ ) follows a normal distribution.

The mean of the the sampling distribution of the sample proportion is: $E(\hat p)=p=0.30$

The standard deviation of the the sampling distribution of the sample proportion is:

$SD(\hat p)=\sqrt{\frac{ p(1- p)}{n} }=\sqrt{\frac{ 0.30(1-0.30)}{25} }=0.092$

(a)

Compute the probability that the number of successes is at most 5 as follows:

The probability of 5 successes is: $p=\frac{5}{25} =0.20$

$P(\hat p\leq 0.20)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq \frac{0.20-0.30}{0.092} )\\=P(Z\leq -1.087)\\=1-P(Z$

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 5 is 0.1379.

The exact probability that the number of successes is at most 5 is:

$P(S\leq 5)={25\choose 5}(0.30)^{5}91-0.30)^{25-5}=0.1935$

The exact probability is more than the approximated probability.

(b)

Compute the probability that the number of successes is at most 7 as follows:

The probability of 5 successes is: $p=\frac{7}{25} =0.28$

$P(\hat p\leq 0.28)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq \frac{0.28-0.30}{0.092} )\\=P(Z\leq -0.2174)\\=1-P(Z$

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 7 is 0.4129.

The exact probability that the number of successes is at most 7 is:

$P(S\leq 57)={25\choose 7}(0.30)^{7}91-0.30)^{25-7}=0.5118$

The exact probability is more than the approximated probability.

(c)

Compute the probability that the number of successes is at most 9 as follows:

The probability of 5 successes is: $p=\frac{9}{25} =0.36$

$P(\hat p\leq 0.36)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq \frac{0.36-0.30}{0.092} )\\=P(Z\leq 0.6522)\\=0.7422$

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 9 is 0.7422.

The exact probability that the number of successes is at most 9 is:

$P(S\leq 9)={25\choose 9}(0.30)^{9}91-0.30)^{25-9}=0.8106$

The exact probability is more than the approximated probability.

(d)

Compute the probability that the number of successes is at most 11 as follows:

The probability of 5 successes is: $p=\frac{11}{25} =0.44$

$P(\hat p\leq 0.44)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq \frac{0.44-0.30}{0.092} )\\=P(Z\leq 1.522)\\=0.9357$

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 11 is 0.9357.

The exact probability that the number of successes is at most 11 is:

$P(S\leq 11)={25\choose 11}(0.30)^{11}91-0.30)^{25-11}=0.9558$

The exact probability is more than the approximated probability.

6 0

3 years ago

The histogram shows the distribution of minting year for a sample of

Anettt [7]

All the parts of the questions about the given histogram have been answered below.

<h3>What is a histogram?</h3>

A histogram is a graph that depicts the frequency of numerical data using rectangles.
The vertical axis of a rectangle represents the distribution frequency of a variable (the amount, or how often that variable appears).

(A) The distribution of minting year:

According to the histogram, the distribution of minting years is 1900-1910 and 1940-2020.

(B) The median:

Add the two middle numbers together and divide them by two to get the median.

10 + 15 / 2 = 12.5

So, the median is somewhere between 10-15.

(C) The range:
The range is computed by subtracting the lowest and highest values.

25 - 0 = 25

So, the range is 25.

(D) The potential outliers:

According to the histogram, the potential outliers are 1910-1940.

Therefore, all the parts of the questions have been answered.

Know more about histograms here:

brainly.com/question/2962546

#SPJ4

8 0

2 years ago

Y 65° 25 can someone solve this ?

n200080 [17]

Answer: a=25

Explanation: I am assuming you are looking for “a” so if you heard the method of 45, 45, 90 it helps you solve it. So by the 45 being across from 25 makes the 45 across a 25 as well, hope this was what you were looking for…

7 0

3 years ago