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4 years ago

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.

2 + 3 = 5 < 7 NOT

21 + 23 = 43 < 44 NOT

12 + 36 = 48 < 53 NOT

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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suppose that the lifetime of a transistor is a gamma random variable x with mean of 24 weeks and standard deviation of 12 weeks.

emmainna [20.7K]

The probability that the transistor will last between 12 and 24 weeks is 0.424

X= lifetime of the transistor in weeks E(X)= 24 weeks

O,= 12 weeks

The anticipated value, variance, and distribution of the random variable X were all provided to us. Finding the parameters alpha and beta is necessary before we can discover the solutions to the difficulties.

X~gamma( $\alpha ,\beta$ )

E(X)= $\alpha \beta$ $\beta$ = $12^{2}/24$ =6 weeks

V(x)= $\alpha \beta ^{2}$ $\alpha$ =24/6= 4

Now we can find the solutions:

The excel formula used to create Figure one is as follows:

=gammadist(X, $\alpha$ , $\beta$ , False)

P( $12\leq X\leq 24$ )

P( $12/6\leq G\leq 24/6$ )

P( $2\leq G\leq 4$ )

P= 0.424

Therefore, probability that the transistor will last between 12 and 24 weeks is 0.424

To learn more about probability click here:

brainly.com/question/11234923

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4 0

1 year ago

Read 2 more answers

4. The following intervals were plausible population proportions for a given sample. Find the margin of error in each case.

Vedmedyk [2.9K]

Answer: a. 0.15 b. 0.03 c. 0.055 d. 0.05

Step-by-step explanation:

For plausible intervals of population proportions , the margin of error is basically half of the difference between upper limit and lower limit.

a. Interval = From 0.35 to 0.65

Upper limit = 0.65

Lower limit = 0.35

Difference = Upper limit- Lower limit

= 0.65-0.35=0.30

Margin of error = (Difference)÷2 =(0.30)÷2=0.15

b. Interval = From 0.72 to 0.78

Upper limit = 0.78

Lower limit = 0.72

Difference = 0.78-0.72 = 0.06

Margin of error =(0.06)÷2=0.03

c. Interval = From 0.84 to 0.95

Upper limit = 0.95

Lower limit = 0.84

Difference =0.95-0.84= 0.11

Margin of error = (0.11)÷2 = 0.055

d.Interval = From 0.47 to 0.57

Upper limit = 0.57

Lower limit = 0.47

Difference = 0.57-0.47=0.10

Margin of error = (0.10)÷2=0.05

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3 years ago

Conrad has a collection of three types of coins: nickels,dimes, and quarters. There are five more nickels than quarters but four

Liono4ka [1.6K]

2 nickles 3 dimes and 4 quarters

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3 years ago

What’s the common difference for this arithmetic sequence

Levart [38]

52 - 37 = 15
67 - 52 = 15
82 - 67 = 15

The common difference for this arithmetic sequence is 15.

Answer: C.

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3 years ago

2/3 - 5bx = bx + 1/3 In the equation shown above, b is a constant. For what value of b = NO SOLUTIONS? A. 5 B. 0 C. -5 D. 2/5

Dmitry_Shevchenko [17]

Answer:

Step-by-step explanation:

Here, we want to know at what values of b does the equation becomes not solvable

Now looking at the left hand side, we have;

2/(3-5bx) and also the right hand side bx + 1/3

For this expression if we insert b = 0, then automatically x cancels out on both sides of the equation and we shall be left with nothing to solve

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3 years ago