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abruzzese [7]
3 years ago
7

If two sides of a triangle are 6 and 8 the third side could be

Mathematics
1 answer:
-Dominant- [34]3 years ago
6 0
3,4,5 are Pythagorean triplets. then 6,8,10 are, so the answer is 10
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In a triangle the sides has 82 degrees ,2x+10 degrees and x degrees what is x
yaroslaw [1]

Answer:

x = 29 1/3 degrees

Step-by-step explanation:

A triangle has 3 angle that total 180 degrees.

The three angles given are 82 degrees ,2x+10 degrees and x degrees.

So,  

82+ (2x+10) + x = 180

Combine like terms.

92 + 3x = 180

Subtract 92 from each side

3x = 180 -92

3x = 88

divide each side by 3

3x/3 = 88/3

Change 88/ 3 to a mixed number

3 goes in to 88 29 times  with 1 left over.

x = 29 1/3 degrees


4 0
3 years ago
Rita discovered that a lab thermometer in her Chemistry class is not accurate. She created the table to show the relationship be
NARA [144]
The correct answer is letter B
6 0
3 years ago
Read 2 more answers
Beth can run 3 miles in 25.8 minutes. How long does it take her to run1 one mile you
Dmitry_Shevchenko [17]

Answer: 8.6

Step-by-step explanation:

25.8 divided by 3

3 0
3 years ago
Read 2 more answers
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 <em>S</em> = a basketball player scores a shot.

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

The number of sample selected is, <em>n</em> = 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
Find the volume of a right circular cone that has a height of 17.6 cm and a base with a
Morgarella [4.7K]

Answer:

1239.3 cm^3

Step-by-step explanation:

8 0
3 years ago
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