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IceJOKER [234]
4 years ago
7

Which of the following side lengths can form a triangle?

Mathematics
2 answers:
daser333 [38]4 years ago
7 0

The sum of the length of the two sides of the triangle must be greater than the length of the third side.


A.

2 + 3 = 5 < 7 NOT


B.

21 + 23 = 43 < 44 NOT


C.

12 + 36 = 48 < 53 NOT


D.

14 + 17 = 31 > 30

14 + 31 = 45 > 17

17 + 31 = 48 > 14

CORRECT


Answer: D. 14 ft, 17 ft, and 30 ft

Margaret [11]4 years ago
3 0

Okay, so my last answer was deleted because I didnt show any work.


According to the rule,


a+b>c

c+b>a

c+a>b


so...


2+3=6...6<7

21+23=44...44+44

12+36=48...48<53

14+17=31...31>30


Since answer D is the only answer that fullfills the rule a+b>c, it is the correct answer


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Answer:

d

Step-by-step explanation:

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What is the value of the expression 5x − y when x = 3 and y = 1?<br> 14<br> 10<br> 7<br> 0
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Let X denote the number of flaws along a 100-m reel of magnetic tape (an integer-valued variable). Suppose Zdenote the number of
Lemur [1.5K]

Answer:

a) P(20 \leq X \leq 30) = P(20-0.5 \leq X \leq 30+0.5)

P(19.5 \leq X \leq 30.5) = P(\frac{19.5-25}{5} \leq Z \leq \frac{30.5 -25}{5})=P(-1.1 \leq Z \leq 1.1)

And we can find this probability like this:

P(-1.1 \leq Z \leq 1.1)= P(Z\leq 1.1) -P(Z\leq -1.1) = 0.864-0.136= 0.728

b) P(X \leq 30)= P(X

And using the z score we got:

P(X

c) P(X

And if we use the continuity correction we got:

P(X

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Continuity correction means that we need to add and subtract 0.5 before standardizing the value specified.

Part a

Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

X \sim N(25,5)  

Where \mu=25 and \sigma=5

Part a

For this case we want to find this probability:

P(20 \leq X \leq 30) = P(20-0.5 \leq X \leq 30+0.5)

And if we use the z score given by:

z =\frac{x-\mu}{\sigma}

We got this:

P(19.5 \leq X \leq 30.5) = P(\frac{19.5-25}{5} \leq Z \leq \frac{30.5 -25}{5})=P(-1.1 \leq Z \leq 1.1)

And we can find this probability like this:

P(-1.1 \leq Z \leq 1.1)= P(Z\leq 1.1) -P(Z\leq -1.1) = 0.864-0.136= 0.728

Part b

For this case we want this probability:

P(X \leq 30)

And if we use the continuity correction we got:

P(X \leq 30)= P(X

And using the z score we got:

P(X

Part c

For this case we want this probability:

P(X

And if we use the continuity correction we got:

P(X

3 0
3 years ago
Please help test due in 1 hour and it’s a grade! Will give good review and brainliest!
jeka94

Answer:

Step-by-step explanation:

(3x-2)+(2x+16)=90

5x=76

x=15.2

I hope this helps

7 0
3 years ago
An article suggests that a poisson process can be used to represent the occurrence of structural loads over time. suppose the me
kirill115 [55]

Answer:

a) \lambda_1 = 2*2 = 4

And let X our random variable who represent the "occurrence of structural loads over time" we know that:

X(2) \sim Poi (4)

And the expected value is E(X) = \lambda =4

So we expect 4 number of loads in the 2 year period.

b) P(X(2) >6) = 1-P(X(2)\leq 6)= 1-[P(X(2) =0)+P(X(2) =1)+P(X(2) =2)+...+P(X(2) =6)]

P(X(2) >6) = 1- [e^{-4}+ \frac{e^{-4}4^1}{1!}+ \frac{e^{-4}4^2}{2!} +\frac{e^{-4}4^3}{3!} +\frac{e^{-4}4^4}{4!}+\frac{e^{-4}4^5}{5!}+\frac{e^{-4}4^6}{6!}]

And we got: P(X(2) >6) =1-0.889=0.111

c)  e^{-2t} \leq 2

We can apply natural log in both sides and we got:

-2t \leq ln(0.2)

If we multiply by -1 both sides of the inequality we have:

2t \geq -ln(0.2)

And if we divide both sides by 2 we got:

t \geq \frac{-ln(0.2)}{2}

t \geq 0.8047

And then we can conclude that the time period with any load would be 0.8047 years.

Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

P(X=x)=\lambda e^{-\lambda x}

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution"

Solution to the problem

Let X our random variable who represent the "occurrence of structural loads over time"

For this case we have the value for the mean given \mu = 0.5 and we can solve for the parameter \lambda like this:

\frac{1}{\lambda} = 0.5

\lambda =2

So then X(t) \sim Poi (\lambda t)

X follows a Poisson process

Part a

For this case since we are interested in the number of loads in a 2 year period the new rate would be given by:

\lambda_1 = 2*2 = 4

And let X our random variable who represent the "occurrence of structural loads over time" we know that:

X(2) \sim Poi (4)

And the expected value is E(X) = \lambda =4

So we expect 4 number of loads in the 2 year period.

Part b

For this case we want the following probability:

P(X(2) >6)

And we can use the complement rule like this

P(X(2) >6) = 1-P(X(2)\leq 6)= 1-[P(X(2) =0)+P(X(2) =1)+P(X(2) =2)+...+P(X(2) =6)]

And we can solve this like this using the masss function:

P(X(2) >6) = 1- [e^{-4}+ \frac{e^{-4}4^1}{1!}+ \frac{e^{-4}4^2}{2!} +\frac{e^{-4}4^3}{3!} +\frac{e^{-4}4^4}{4!}+\frac{e^{-4}4^5}{5!}+\frac{e^{-4}4^6}{6!}]

And we got: P(X(2) >6) =1-0.889=0.111

Part c

For this case we know that the arrival time follows an exponential distribution and let T the random variable:

T \sim Exp(\lambda=2)

The probability of no arrival during a period of duration t is given by:

f(T) = e^{-\lambda t}

And we want to find a value of t who satisfy this:

e^{-2t} \leq 2

We can apply natural log in both sides and we got:

-2t \leq ln(0.2)

If we multiply by -1 both sides of the inequality we have:

2t \geq -ln(0.2)

And if we divide both sides by 2 we got:

t \geq \frac{-ln(0.2)}{2}

t \geq 0.8047

And then we can conclude that the time period with any load would be 0.8047 years.

3 0
4 years ago
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