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3 years ago

Can a triangle be made using the side lengths of 8, 2, 10?

Mathematics

1 answer:

Bas_tet [7]3 years ago

4 0

the triangle can't be formed.....

hence, correct answer is C. No, the sum of the shortest sides measure is equal to the longest side.

You might be interested in

The volume of a cube is 216mm3.What is the length of its side?

Studentka2010 [4]

Steps:
216 mm3 is the volume.

Given the volume, you can take the cube root of the volume to find the side.
The cube root of 216 is 6.

Answer: 6

8 0

3 years ago

The probability distribution for the rate of return on an investment is

babymother [125]

Answer:

a)0.7

b) 10.03

c) 0.0801

Step-by-step explanation:

Rate of return Probability

9.5 0.1

9.8 0.2

10 0.3

10.2 0.3

10.6 0.1

P(Rate of return is at least 10%)=P(R=10)+P(R=10.2)+P(R=10.6)

P(Rate of return is at least 10%)=0.3+0.3+0.1

P(Rate of return is at least 10%)=0.7

Expected rate of return=E(x)=sum(x*p(x))

Rate of return(x) Probability(p(x)) x*p(x)

9.5 0.1 0.95

9.8 0.2 1.96

10 0.3 3

10.2 0.3 3.06

10.6 0.1 1.06

Expected rate of return=E(x)=sum(x*p(x))

Expected rate of return=0.95+1.96+3+3.06+1.06=10.03

variance of the rate of return=V(x)= $sum(x^2p(x))-[sum(x*p(x))]^2$

Rate of return(x) Probability(p(x)) x*p(x) x²*p(x)

9.5 0.1 0.95 9.025

9.8 0.2 1.96 19.208

10 0.3 3 30

10.2 0.3 3.06 31.212

10.6 0.1 1.06 11.236

sum[x²*p(x)]=9.025+19.208+30+31.212+11.236=100.681

variance of the rate of return=V(x)=sum(x²*p(x))-[sum(x*p(x))]²

variance of the rate of return=V(x)=100.681-(10.03)²

variance of the rate of return=V(x)=100.681-100.6009

variance of the rate of return=V(x)=0.0801

6 0

3 years ago

. Exam scores for a large introductory statistics class follow an approximate normal distribution with an average score of 56 an

Andru [333]

Answer:

0.1% probability that the average score of a random sample of 20 students exceeds 59.5.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$

Normal probability distribution

Problems of normally distributed samples can be 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.

In this problem, we have that:

$\mu = 56, \sigma = 5, n = 20, s = \frac{5}{\sqrt{20}} = 1.12$