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Svetradugi [14.3K]
3 years ago
11

Quando existe, dizemos que a solução de um sistema de equações lineares é o conjunto ordenado de números que satisfaz todas as e

quações que o compõe. Você foi sorteado para participar de um show de prêmios em um programa de televisão da sua cidade. A regra do jogo é a seguinte: para cada resposta certa, você ganha R$ 200,00; para cada resposta errada, perde R$ 150,00. No início do jogo, você recebeu R$ 500,00 em créditos. Se você respondeu todas as 25 perguntas e terminou o jogo com R$ 600,00, quantas perguntas errou?
Mathematics
1 answer:
QveST [7]3 years ago
5 0
X representa o numero das suas respostas certas.
y represnta o numero das suas respostas erradas.
O total de perguntas <span>é 25, portanto
x + y = 25

Agora tratamos do dinheiro.
Come</span>ça com <span>R$ 500,00
Pelas x respostas certas, recebe 200x.
Pelas y respostas errads perde 150y.
O total de dineheiro inicial mais os ganhos menos as perdas s</span>ão iguais a
R$ 600,00, portanto
500 + 200x - 150y = 600
200x - 150y = 100
20x - 15y = 10

Temos um sistema de duas equações com duas variaveis.

<span>x + y = 25</span>
20x - 15y = 10

      15x + 15y = 375
+    20x - 15y = 10
---------------------------
       35x         = 385
           x         = 11

x + y = 25

11 + y = 25

y = 14

Resposta: Errou 14 perguntas.


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Answer:

a) 0.5.

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Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 6, \sigma = 0.2

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This is 1 subtracted by the pvalue of Z when X = 6. So

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Z = \frac{6-6}{0.2}

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Z = 0 has a pvalue of 0.5.

1 - 0.5 = 0.5.

(b) P(x < 6.2)=

This is the pvalue of Z when X = 6.2. So

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Z = \frac{6.2-6}{0.2}

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(c) P(x ≤ 6.2) =

In the normal distribution, the probability of an exact value, for example, P(X = 6.2), is always zero, which means that P(x ≤ 6.2) = P(x < 6.2) = 0.8413.

(d) P(5.8 < x < 6.2) =

This is the pvalue of Z when X = 6.2 subtracted by the pvalue of Z when X  5.8.

X = 6.2

Z = \frac{X - \mu}{\sigma}

Z = \frac{6.2-6}{0.2}

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X = 5.8

Z = \frac{X - \mu}{\sigma}

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Z = -1

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(e) P(x > 5.7) =

This is 1 subtracted by the pvalue of Z when X = 5.7.

Z = \frac{X - \mu}{\sigma}

Z = \frac{5.8-6}{0.2}

Z = -1.5

Z = -1.5 has a pvalue of 0.0668

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(f) P(x > 5) =

This is 1 subtracted by the pvalue of Z when X = 5.

Z = \frac{X - \mu}{\sigma}

Z = \frac{5-6}{0.2}

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