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3 years ago

Is the triangle with these sides a right triangle? 22mm, 26mm, 34 mm

2 answers:

7 0

If a^2 + b^2 = c^2, where c is always the longest side, then we have a right triangle.
a = 22
b = 26
c = 34

Plug them into the pythagorean theorem formula
a^2 + b^2 = c^2
22^2 + 26^2 = 34^2
484 + 676 = 1156
1160 = 1156

The last equation is false, so the original equation is false for that given (a,b,c) triple

These sides do NOT form a right triangle.

Liono4ka [1.6K]3 years ago

3 0

If a^2 + b^2 = c^2 then it is a right triangle.
22^2 + 26^2 ?=? 34^2
484 + 676 ?=? 1156
1160 ?=? 1156
This is NOT TRUE so it is not a right triangle.

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Which expression is represented by the diagram? o -6-(-1) o -6-1 o -5-(-1) o -5-1

cupoosta [38]

Option A:

–6 – (–1)

Solution:

Minus symbol are given in the diagram.

There are 6 minus symbol in the diagram.

From that expression 1 minus symbol is cancelled.

To find which expression represents this situation:

Option A: –6 – (–1)

Total number of minus = –6

From this cancel one minus: –6 – (–1)

–6 – (–1) = –6 + 1 = –5

So, remaining minus are –5.

This is the correct answer.

Option B: –6 –1

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But note that remaining 5 minus are only there in the diagram.

So, this is not correct answer.

Option C: –5 – (–1)

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So, this is not correct answer.

Option D: –5 –1

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So, this is not correct answer.

Hence –6 – (–1) is the correct expression which represented the diagram.

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3 years ago

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Multiply 60 times 7 410, 420, 67, or 430?

DochEvi [55]

Answer:

420

Step-by-step explanation:

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3 years ago

he amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.3 minutes and s

sp2606 [1]

Complete question:

He amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.3 minutes and standard deviation 1.4 minutes. Suppose that a random sample of n equals 47 customers is observed. Find the probability that the average time waiting in line for these customers is

a) less than 8 minutes

b) between 8 and 9 minutes

c) less than 7.5 minutes

Answer:

a) 0.0708

b) 0.9291

c) 0.0000

Step-by-step explanation:

Given:

n = 47

u = 8.3 mins

s.d = 1.4 mins

a) Less than 8 minutes:

$P(X$

P(X' < 8) = P(Z< - 1.47)

Using the normal distribution table:

NORMSDIST(-1.47)

= 0.0708

b) between 8 and 9 minutes:

P(8< X' <9) = $[\frac{8-8.3}{1.4/ \sqrt{47}}< \frac{X'-u}{s.d/ \sqrt{n}} < \frac{9-8.3}{1.4/ \sqrt{47}}]$

= P(-1.47 <Z< 6.366)

= P( Z< 6.366) - P(Z< -1.47)

Using normal distribution table,

$NORMSDIST(6.366)-NORMSDIST(-1.47)$

0.9999 - 0.0708

= 0.9291

c) Less than 7.5 minutes:

P(X'<7.5) = $P [Z< \frac{7.5-8.3}{1.4/ \sqrt{47}}]$

P(X' < 7.5) = P(Z< -3.92)

NORMSDIST (-3.92)

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3 years ago

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Answer:

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The answer would have came out of 3.84615384615

Then round up the answer to 3.85. So the unit price is $3.85.

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