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3 years ago

Help please!!!!!!!!!

Mathematics

1 answer:

SOVA2 [1]3 years ago

6 0

Find the area of the triangles and smal ler square and add them together ; and find the area of the complete square and set them equal to each other (these are known truths)
triangle: l×w /2
square: s^2
There are 4 triangles so

4 (1/2ab)+c^2= (a+b)(a+b)
simplify

2ab+c^2= a^2+2ab+b^2
-2ab both sides

c^2=a^2+b^2

tada!! it's proven :)

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The price of a table at Best Buy is $150. If the sales tax is 6%, what is the final price of the table including tax?

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Answer:

159

Step-by-step explanation:

100%-> 150

106%->159

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Solve the given system, using the substitution method.

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Y = 3x - 7 . . . . . (1)
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A real estate agent has 8 master keys to open several new homes. Only 1 master key will open any given house. If 40% of these ho

riadik2000 [5.3K]

Answer:

Step-by-step explanation:

Suppose F represents the value that a room is unlocked.

Then, we can assume that the probability of unlocked homes is:

P(unlocked homes) = P(U)

P(unlocked homes) = 40%

P(unlocked homes) = 0.40

Also, let us represent the value of the locked room with G.

P(locked homes) = P(G)

P(locked homes) = (1 - 0.40)

P(locked homes) = 0.60

Let the probability of selecting a correct key be P(S)

It implies that for the agent to use 3 keys, we have a combination of $^8C_3$ possible ways for the set of keys.

Now; since only one will open the house, then:

P(select correct key) = P(S)

$P(S) = \dfrac{ (^1_1) (^7_2) }{ ^8_3 }$

$P(S) = \dfrac{21}{56}$

P(S) = 0.375

Finally, for the real estate agent to have access to specific homes supposing the agent select three master keys at random prior to the time he left his office, Then:

P(F ∪ (G∩S) = P(F) + P(G∩S)

P(F ∪ (G∩S) = P(F) + P(G) × P(S)

P(F ∪ (G∩S) = 0.40 + (0.60×0.375)

P(F ∪ (G∩S) = 0.40 + 0.225

P(F ∪ (G∩S) =0.625

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3 years ago

A manufacturer of chocolate chips would like to know whether its bag filling machine works correctly at the 418 gram setting. It

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Answer: There is sufficient evidence to support the claim that the bags are under-filled.

Step-by-step explanation:

Since we have given that

$H_0:\mu=418\\\\H_a:\mu$

Sample mean = 411

Standard deviation = 20

n = 9

So, the test statistic value is given by

$z=\dfrac{\bar{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}\\\\\\z=\dfrac{411-418}{\dfrac{20}{\sqrt{9}}}\\\\\\z=\dfrac{-7}{\dfrac{20}{3}}\\\\\\z=-1.05$

At 0.025 level of significance,

critical value z = -2.306

since -2.306<-1.05

so, we will reject the null hypothesis.

Yes, there is sufficient evidence to support the claim that the bags are underfilled.

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Answer:

All of them can be arranged.

Step-by-step explanation:

Hope this helps. :D

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