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Mandarinka [93]
3 years ago
5

Ami has a mass of 72 kg.

Mathematics
1 answer:
il63 [147K]3 years ago
4 0

{\huge{\underline{\sf{Answer}}}}

<h3>» Ami's mass is \sf \frac{2}{3} the mass of Raphael.</h3>

<u>S</u><u>o</u><u>l</u><u>u</u><u>t</u><u>i</u><u>o</u><u>n</u><u>:</u>

Given that

  • Ami's mass is 72 kg and
  • Raphael's mass is 108 kg.

To write Ami’s mass as a fraction of Raphael's mass, we divide Ami's mass by Raphael's mass. So it is now \sf \frac{72}{108}.

However, there the fraction is not yet in its simplest form, so we have to extract common factors from both the numerator and the denominator:

  • \sf \frac{36 \times 2}{36 \times 3}

  • = {\boxed{\sf \frac{2}{ 3}}}

Therefore, Ami's mass is \sf \frac{2}{3} the mass of Raphael.

{\: \:}

{\huge{\underline{\sf{Hope\:It\:Helps}}}}

#PrideOfThePhilippines

#LetsLearn #BeBrainly

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<em />

Step-by-step explanation:

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<em>What should I buy? A study conducted by a research group in a recent year reported that 57% of cell phone owners used their phones inside a store for guidance on purchasing decisions. A sample of 14 cell phone owners is studied. Round the answers to at least four decimal places. What is the probability that seven or more of them used their phones for guidance on purchasing decisions? </em>

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P(x=k)=\dbinom{n}{k} p^{k}q^{n-k}

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P(x\geq7)=\sum_{k=7}^{14}P(x=k)\\\\\\

P(x=7)=\dbinom{14}{7} p^{7}q^{7}=3432*0.0195*0.0027=0.1824\\\\\\P(x=8) = \dbinom{14}{8} p^{8}q^{6}=3003*0.0111*0.0063=0.2115\\\\\\P(x=9) = \dbinom{14}{9} p^{9}q^{5}=2002*0.0064*0.0147=0.1869\\\\\\P(x=10) = \dbinom{14}{10} p^{10}q^{4}=1001*0.0036*0.0342=0.1239\\\\\\P(x=11) = \dbinom{14}{11} p^{11}q^{3}=364*0.0021*0.0795=0.0597\\\\\\P(x=12) = \dbinom{14}{12} p^{12}q^{2}=91*0.0012*0.1849=0.0198\\\\\\P(x=13) = \dbinom{14}{13} p^{13}q^{1}=14*0.0007*0.43=0.004\\\\\\

P(x=14) = \dbinom{14}{14} p^{14}q^{0}=1*0.0004*1=0.0004\\\\\\

P(x\geq7)=0.1824+0.2115+0.1869+0.1239+0.0597+0.0198+0.004+0.0004\\\\P(x\geq7)=0.7886

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