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kozerog [31]
2 years ago
15

Help me I don’t understand it?

Mathematics
1 answer:
photoshop1234 [79]2 years ago
3 0

let's do 1, 5, 6, 10 and 13.

1)

well, the denominator is the same on each, so we simply have to look at the numerator, who is larger 3 or 5?  3 < 5, 3 is less than 5, so then........\bf \cfrac{3}{12}

5)

\bf \begin{cases} \cfrac{2}{4}\implies \cfrac{1}{2} \\\\\\ \cfrac{10}{20}\implies \cfrac{1}{2} \end{cases}\qquad \implies \cfrac{2}{4}=\cfrac{10}{20}\implies \cfrac{1}{2}=\cfrac{1}{2}

6)

we can make both denominators the same if we simply <u>multiply each fraction by the other's denominator</u>.


\bf \begin{cases} \cfrac{1}{4}\cdot \cfrac{13}{13}\implies \cfrac{13}{52} \\\\\\ \cfrac{4}{13}\cdot \cfrac{4}{4}\implies \cfrac{16}{52} \end{cases}\qquad \implies \cfrac{13}{52}

10)

we'll convert the mixed fractions to improper fractions first, then make their denominator the same just like we did in 6).


\bf \stackrel{mixed}{4\frac{1}{7}}\implies \cfrac{4\cdot 7+1}{7}\implies \stackrel{improper}{\cfrac{29}{7}}~\hfill \stackrel{mixed}{3\frac{3}{18}}\implies \cfrac{3\cdot 18+3}{18}\implies \stackrel{improper}{\cfrac{57}{18}} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} \cfrac{29}{7}\cdot \cfrac{18}{18}\implies \cfrac{522}{126} \\\\\\ \cfrac{57}{18}\cdot \cfrac{7}{7}\implies \cfrac{399}{126} \end{cases}\qquad \implies \cfrac{522}{126}>\cfrac{399}{126}\qquad therefore\qquad 4\frac{1}{7}>3\frac{3}{18}


13)

so both fractions are at a value from 9, so we can simply say, which is larger 2/6 or 4/12?


\bf \begin{cases} \cfrac{2}{6}\implies \cfrac{1}{3} \\\\\\ \cfrac{4}{12}\implies \cfrac{1}{3} \end{cases}\qquad \implies \cfrac{1}{3}=\cfrac{1}{3}\qquad therefore\qquad 9\frac{2}{6}=9\frac{4}{12}

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Step-by-step explanation:

x+128=180

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Suppose that 80% of all trucks undergoing a brake inspection at a certain inspection facility pass the inspection. Consider grou
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Answer:

P=0.147

Step-by-step explanation:

As we know 80% of the trucks have good brakes. That means that probability the 1 randomly selected truck has good brakes is P(good brakes)=0.8 . So the probability that 1 randomly selected truck has bad brakes Q(bad brakes)=1-0.8-0.2

We have to find the probability, that at least 9 trucks from 16 have good brakes, however fewer than 12 trucks from 16 have good brakes. That actually means the the number of trucks with good brakes has to be 9, 10 or 11 trucks from 16.

We have to find the probability of each event (9, 10 or 11 trucks from 16 will pass the inspection) .  To find the required probability 3 mentioned probabilitie have to be summarized.

So P(9/16 )=  C16 9 * P(good brakes)^9*Q(bad brakes)^7

P(9/16 )= 16!/9!/7!*0.8^9*0.2^7= 11*13*5*16*0.8^9*0.2^7=approx 0.02

P(10/16)=16!/10!/6!*0.8^10*0.2^6=11*13*7*0.8^10*0.2^6=approx 0.007

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7 0
2 years ago
I need help please and thank you.
kogti [31]

Step-by-step explanation:

\frac{2x}{5}  +  \frac{1}{3}  =  \frac{7x - 2}{15}

\frac{6x + 5}{15}  =  \frac{7x - 2}{15}

6x + 5 = 7x - 2

6x - 7x =  - 2 - 5

- x =  - 7

x = 7

option (4)

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