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umka21 [38]
3 years ago
14

A ball is thrown straight up with a initial velocity of 56 feet per second . The height , h , of the ball t seconds after it is

thrown is given by the formula h(t)=56t-16t^2 . What is the maximum height of the ball?
Mathematics
1 answer:
Tju [1.3M]3 years ago
3 0

<span>Formula: H(t) = 56t – 16t^2   </span>

H(t) = - 16t^2 + 56t

<span><span>A.      </span>What is the height of the ball after 1 second? H (1) = 56(1) – 16(1) ^2 = 40 pt.</span>

<span><span>B.      </span>What is the maximum height? X = - (56)/2(- 16) = 1.75 sec h (1.75) = 56(1.75) – 16(1.75) ^2 h (1.75) = 49ft.</span>

<span><span>C.      </span><span>After how many seconds will it return to the ground? – 16t^2 + 56t = 0   - 8t =0 t = 0</span></span>

<span><span>-          </span><span>8t (2 + - 7) = 0     2t – 7 = 0  t = 7/2   Ans: 3.5 seconds</span></span>

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Step-by-step explanation:

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Step-by-step explanation:

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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.

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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\\\\\\

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