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Svetradugi [14.3K]
4 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]4 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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(9.885–0.365)÷1.7+4.4
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Answer:

The answer is 10

Step-by-step explanation:

BODMAS:

(9.885-0.365)÷1.7+4.4

= 9.52 ÷ 1.7 + 4.4

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In a school, the ratio of musicians to athletes is 3 : 1. The number of musicians is how many times the number of athletes?  ANS
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The number of musicians in the school is 3 times the number of athletes

Step-by-step explanation:


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Construc t a 95% confidence interval for the population standard deviation σ of a random sample of 15 men who have a mean weight
GrogVix [38]

The 95% confidence interval for the population standard deviation will be,

$\sqrt{\frac{(n-1) s^{2}}{\chi^{2}} \frac{\alpha}{2}} & < \sigma < \sqrt{\frac{(n-1) s^{2}}{\chi^{2}}} \\

simplifying the equation, we get

$\sqrt{\frac{2551.5}{26.1}} & < \sigma < \sqrt{\frac{2551.5}{5.63}} \\9.887 & < \sigma < 21.288

The 95% confidence interval exists (9.887, 21.288).

<h3>What is the  95% of confidence interval?</h3>

A random sample of 15 men exists selected. The mean weight exists at $165.2 pounds and the standard deviation exists at $13.5 pounds. The population exists normally distributed.

So, $n=15, \bar{X}=165.2, s=13.5$

where n exists the sample size, $\bar{X}$ exists the sample size

s exists the sample standard deviation.

The degrees of freedom will be n - 1 i.e.15 - 1 = 14.

For a 95% confidence interval, the level of significance will be $\alpha=0.05$.

The 95% confidence interval for the population standard deviation will be,

$\sqrt{\frac{(n-1) s^{2}}{\chi^{2}} \frac{\alpha}{2}} & < \sigma < \sqrt{\frac{(n-1) s^{2}}{\chi^{2}}} \\

substitute the values in the above equation, we get

$\sqrt{\frac{(15-1)(13.5)^{2}}{\chi^{2}} 0.05} & < \sigma < \sqrt{\frac{(15-1)(13.5)^{2}}{\chi_{0}^{2}}} \\

$\sqrt{\frac{(15-1)(13.5)^{2}}{\chi_{0.975}^{2}}} & < \sigma < \sqrt{\frac{(15-1)(13.5)^{2}}{\chi_{0.025}^{2}}} \\

simplifying the above equation, we get

$\sqrt{\frac{2551.5}{26.1}} & < \sigma < \sqrt{\frac{2551.5}{5.63}} \\9.887 & < \sigma < 21.288

The 95% confidence interval exists (9.887, 21.288).

To learn more about confidence interval refer to:

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Step-by-step explanation:

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3 years ago
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Answer:

4:1

Step-by-step explanation:

Because AB is the opposite of BA, you flip the fraction, to get 4/1, or 4:1

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