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lisabon 2012 [21]
3 years ago
5

Connor spent $66 on rookie cards and Hall of Fame cards how many of each type of car did he buy rookie cards are four for six Ha

ll of Fame two for nine
Mathematics
1 answer:
barxatty [35]3 years ago
5 0

Answer:

We conclude that he purchased 20  rookie cards and 8  Hall of Fame cards.

Step-by-step explanation:

We know that Connor spent $66 on rookie cards and Hall of Fame cards. We know that rookie cards are four for six Hall of Fame two for nine.

Therefore, we calculate:

5\cdot 6 +4\cdot 9=30+36=66

5·4= 20 rookie cards

4·2= 8 Hall of Fame cards

We conclude that he purchased 20  rookie cards and 8  Hall of Fame cards.

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Answer:

(a1) The probability that temperature increase will be less than 20°C is 0.667.

(a2) The probability that temperature increase will be between 20°C and 22°C is 0.133.

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Step-by-step explanation:

Let <em>X</em> = temperature increase.

The random variable <em>X</em> follows a continuous Uniform distribution, distributed over the range [10°C, 25°C].

The probability density function of <em>X</em> is:

f(X)=\left \{ {{\frac{1}{25-10}=\frac{1}{15};\ x\in [10, 25]} \atop {0;\ otherwise}} \right.

(a1)

Compute the probability that temperature increase will be less than 20°C as follows:

P(X

Thus, the probability that temperature increase will be less than 20°C is 0.667.

(a2)

Compute the probability that temperature increase will be between 20°C and 22°C as follows:

P(20

Thus, the probability that temperature increase will be between 20°C and 22°C is 0.133.

(b)

Compute the probability that at any point of time the temperature increase is potentially dangerous as follows:

P(X>18)=\int\limits^{25}_{18}{\frac{1}{15}}\, dx\\=\frac{1}{15}\int\limits^{25}_{18}{dx}\,\\=\frac{1}{15}[x]^{25}_{18}=\frac{1}{15}[25-18]=\frac{7}{15}\\=0.467

Thus, the probability that at any point of time the temperature increase is potentially dangerous is 0.467.

(c)

Compute the expected value of the uniform random variable <em>X</em> as follows:

E(X)=\frac{1}{2}[10+25]=\frac{35}{2}=17.5

Thus, the expected value of the temperature increase is 17.5°C.

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