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RoseWind [281]
3 years ago
7

A rectangular block of metal is 256mm long, 152mm wide and 81mm high. If the metal block is melted to form a cube, find the leng

th of each side of the cube.
(answer: 147mm)
please explain with details
Mathematics
1 answer:
Nimfa-mama [501]3 years ago
8 0

Answer:

147mm

Step-by-step explanation:

volume of the block:

256×152×81

3,151,872 mm³

volume of a cube: side³

side³ = 3,151,872

side = 3,151,872^⅓

side = 146.618753863

approximately 147 mm

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Answer:


Step-by-step explanation:

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4. divide entire equation by 2 to isolate y

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6. multiply entire equation by -1 to make it positive

7. y=x+3

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Read 2 more answers
An analysis of sales records for the last 120 weeks gives the following results. Assuming that these past data are a reliable gu
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Answer:

(a) 0.5333

(b) 0.6583

(c) 0.5583

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(e) 0.6167

Step-by-step explanation:

Denote the events as follows:

<em>A</em> = a competitor will advertise

<em>NA </em>= a competitor will not advertise

<em>L </em>= Low sales will be achieved

<em>M </em>= Medium sales will be achieved

<em>H </em>= High sales will be achieved

The data provided is of the form:

          Low (L)    Medium (M)    High (H)    Total    

A            32                   14                 18               64

NA          21                   12                 23              56

Total      53                  26                 41              120

The probability of an event <em>E</em> is:

P(E)=\frac{n(E)}{N}

n (E) = favorable outcomes of event <em>E</em>

N = Total number of outcomes

(a)

Compute the probability that next week the competitor will advertise as follows:

P(A)=\frac{n(A)}{N}=\frac{64}{120}=0.5333

Thus, the probability that next week the competitor will advertise is 0.5333.

(b)

Compute the probability that next week sales will not be high as follows:

P(H^{c})=1-P(H)=1-\frac{n(H)}{N}=1-\frac{41}{120}=\frac{120-41}{120}=0.6583

Thus, the probability that next week sales will not be high is 0.6583.

(c)

The events of achieving a medium or high sales are mutually exclusive.

Since the sales achieved will either be medium or high. They cannot be both.

So, P (M ∩ H) = 0.

Compute the probability that next week there will be medium or high sales will be achieved as follows:

P(M\cup H)=P(M)+P(H)=\frac{26}{120}+\frac{41}{120}=\frac{26+41}{120}=0.5583

Thus, the probability that next week there will be medium or high sales will be achieved is 0.5583.

(d)

Compute the probability that next week either the competitor will advertise, or only low sales will be achieved as follows:

P(A\cup L)=P(A)+P(L)-P(A\cap L)=\frac{64}{120}+\frac{53}{120}-\frac{32}{120}=\frac{85}{120}=0.7083

Thus, the  the probability that next week either the competitor will advertise, or only low sales will be achieved is 0.7083.

(e)

Compute the probability that next week either the competitor will not advertise, or high sales will be achieved as follows:

P(NA\cup H)=P(NA)+P(H)-P(NA\cap H)=\frac{56}{120}+\frac{41}{120}-\frac{23}{120}=\frac{74}{120}=0.6167

Thus, the  the probability that next week either the competitor will not advertise, or high sales will be achieved is 0.6167.

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Answer:

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<em>                           </em>x=\frac{-3\pm \sqrt{3^2-4\cdot \:1\cdot \left(-4\right)}}{2\cdot \:1}

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Third Picture ⇒ Missing Info

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<em>-------------------------------------------------------------------------------------------------------------</em>

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<em>-------------------------------------------------------------------------------------------------------------</em>

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<em>-------------------------------------------------------------------------------------------------------------</em>

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x_1=\frac{-11+1}{2\cdot \:2},\:x_2=\frac{-11-1}{2\cdot \:2}

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<em>-------------------------------------------------------------------------------------------------------------</em>

<em>Suppose you are solving a quadratic equation and this is your work so far...</em>

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<em>Thus, we have:</em>

<em />x=\frac{-\left(-5\right)\pm \sqrt{\left(-5\right)^2-4\cdot \:1\cdot \:6}}{2\cdot \:1}

x=\frac{-\left(-5\right)\pm \:1}{2\cdot \:1}

x=\frac{5\pm\sqrt{1} }{2}

x_1=\frac{-\left(-5\right)+1}{2\cdot \:1},\:x_2=\frac{-\left(-5\right)-1}{2\cdot \:1}

x=3,x=2

<em>-------------------------------------------------------------------------------------------------------------</em>

<u><em>Kavinsky</em></u>

6 0
2 years ago
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