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2 years ago

Can someone please help me with this question.

Mathematics

1 answer:

Ber [7]2 years ago

4 0

Answer:

square root of 2

Step-by-step explanation:

Sides are 1 and 1 and the hypotenuse is the longest side (opposite the right angle)

We can use pythagerouses theorem : a^2+b^2=c^2

this means that 1^2+1^2=c^2

1+1=c^2

2=c^2

c=2/sqrt

c is the square root of 2

the hypotenuse is the square root of 2

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Paladinen [302]

Answer:

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Step-by-step explanation:

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2 years ago

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Tickets to a play cost $5 at the door and $4 in advance. The theater club wants to raise at least $400 from the play. Write and

cestrela7 [59]

Your equation should be
$4(40 tickets) + $5(X) = $400
160 +5x = 400
-160 -160
0 5x = 340
divide both sides by 5
5x/5 = 340/5
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so they need to sell 68 tickets at the door.

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jolli1 [7]

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3 years ago

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What is the answer for 4x y = -10

CaHeK987 [17]

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2 years ago

A shipment of 11 printers contains 2 that are defective. Find the probability that a sample of size 2, drawn from the 11, will

svet-max [94.6K]

The required probability is $\frac{36}{55}$

<u>Solution:</u>

Given, a shipment of 11 printers contains 2 that are defective.

We have to find the probability that a sample of size 2, drawn from the 11, will not contain a defective printer.

Now, we know that, $\text { probability }=\frac{\text { favourable outcomes }}{\text { total outcomes }}$

Probability for first draw to be non-defective $=\frac{11-2}{11}=\frac{9}{11}$

(total printers = 11; total defective printers = 2)

Probability for second draw to be non defective $=\frac{10-2}{10}=\frac{8}{10}=\frac{4}{5}$

(printers after first slot = 10; total defective printers = 2)

Then, total probability $=\frac{9}{11} \times \frac{4}{5}=\frac{36}{55}$

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3 years ago