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Readme [11.4K]
3 years ago
7

Help asap, please!!..

Mathematics
1 answer:
Darya [45]3 years ago
7 0

that alot of work no cap

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What should I buy? A study conducted by a research group in a recent year reported that of cell phone owners used their phones i
Llana [10]

Answer:

The probability that seven or more of them used their phones for guidance on purchasing decisions is 0.7886.

<em />

Step-by-step explanation:

<em>The question is incomplete:</em>

<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>

We can model this as a binomial random variable, with p=0.57 and n=14.

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

a) We have to calculate the probability that seven or more of them used their phones for guidance on purchasing decisions:

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

7 0
3 years ago
A small home speaker produces 0.200 ~\text{W}0.200 W of acoustical power. (This is achieved, for example, by a 10-Watt speaker o
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Answer:

i dont speak english

Step-by-step explanation:

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Megan is trying to break the school's long jump
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Find the area of the region shaded in green. Use 3.14 to approximate pi.
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Answer:

254 cm² (to 3 s.f.)

Step-by-step explanation:

Area of shaded region

= area of large circle -area of smaller circle

\boxed{area \: of \: circle = \pi {r}^{2} }

Radius of large circle= 15cm

Radius of smaller circle= 12cm

Area of shaded region

= π(15²) -π(12²)

= 225π -144π

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