Unscramble the words

nikitadnepr [17]

Answer:

5 people

Step-by-step explanation:

For each people Ms Hernandez bring to the zoo, she will pay $15.50, so if she go alone, 1×15.50, if she go with one person, 2×15.50, with three 3×15.50, and keep growing this way. The price each person pay is constant and equal to 15.50, and what will determine the final price is the number of people. Also remember that she always will have to pay $ 10 on parking, so you can write an equation with this:

15.50x +10 = y, as x being the number of people and y being the final price.

She have $100, so this is the max she can spend. Two know the number of people she can bring to the zoo, put 100 in place of y and find the value of x:

15.50x + 10 = 100

15.50x = 100 - 10

15.50x = 90

x = 90/15.50

x = 5.8

But there's no way to bring 0.8 person, so the max she can bring are 5 people, including her

5 0

2 years ago

At an ocean-side nuclear power plant, seawater is used as part of the cooling system. This raises the temperature of the water t

grandymaker [24]

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.

(b) The probability that at any point of time the temperature increase is potentially dangerous is 0.467.

(c) The expected value of the temperature increase is 17.5°C.

Step-by-step explanation:

Let X = temperature increase.

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

The probability density function of X 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:

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