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levacccp [35]
3 years ago
14

Suppose? 30% of the people in your town talk on their cell phones and drive at the same time. if you stood on a busy street corn

er and watched 50 people drive? by, about what proportion would you expect to see talking on their cell? phones?
Mathematics
2 answers:
kirill [66]3 years ago
8 0
30% of 50 = 15 people
Rufina [12.5K]3 years ago
3 0
15 people would be talking on there cell phones
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What are the solutions to the equation 3(x – 4)(x + 5) = 0?
iris [78.8K]

Answer:

x=4 and x= - 5

Step-by-step explanation:

In order to solve this we should re-arrange the equation:

(x-4)(x+5)=\frac{0}{3}

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(x-4)(x+5)=0

Then we can seperate this into:

(x-4)=0\\(x+5)=0

So solving for x in both cases we get

x-4=0 \\x=4\\And\\x+5=0\\x=-5\\

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159/day

Step-by-step explanation:

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15-3(2+6 times -3) <br> What would this simplify to?
pickupchik [31]
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A certain component is critical to the operation of an electrical system and must be replaced immediately upon failure. If the m
nekit [7.7K]

Answer:

Step-by-step explanation:

From the given information:

Assuming we have an integer c to represent the components in the stocks.

Thus, the needed probability can be expressed as:

P(X_1+X_2+ ........ X_c \ge 2000)

To break this down, we have:

P(\sum X_i \ge 2000) = P \Bigg(\dfrac{\sum X_i - 100n}{\sqrt{n \ Var (X)}} \ge \dfrac{2000-100 n }{\sqrt{n \times 900}} \Bigg )

P \Bigg(\dfrac{\sum X_i - 100n}{\sqrt{n \ Var (X)}} \ge \dfrac{2000-100 n }{\sqrt{n \times 900}} \Bigg )= P (Z \ge 0.95 ) \ \ \  because  \ Z_{0.05} = -1.65 \\ \\  \\  \dfrac{2000-100 n }{\sqrt{n \times 900}} = -1.65 \\ \\ \\ \dfrac{100\  n-2000  }{\sqrt{900n}} = 1.65  \\ \\ 100 n -1.65 \sqrt{900 n }-2000 = 0

By solving the equation:

n = 23

Thus, relating to the needed condition; n ≥ 23

The needed number of the components that should be in stock should be  at least 23.

5 0
3 years ago
A fair four-sided die has two faces numbered 0 and two faces numbered 2. Another fair four-sided die has its faces numbered 0, 1
Mandarinka [93]

The two dice have PMFs

P(X=x)=\begin{cases}\dfrac12&\text{for }x\in\{0,2\}\\\\0&\text{otherwise}\end{cases}

P(Y=y)=\begin{cases}\dfrac14&\text{for }y\in\{0,1,4,5\}\\\\0&\text{otherwise}\end{cases}

The sum of the two rolls can take on 8 different values:

W=\begin{cases}0&\text{if }x=0,y=0\\1&\text{if }x=0,y=1\\2&\text{if }x=2,y=0\\3&\text{if }x=2,y=1\\4&\text{if }x=0,y=4\\5&\text{if }x=0,y=5\\6&\text{if }x=2,y=4\\7&\text{if }x=2,y=5\end{cases}

The outcome of either roll is independent of the other, so that

P(X=x\text{ and }Y=y)=P(X=x)P(Y=y)

a. Then PMF of W is

P(W=w)=\begin{cases}\dfrac12\cdot\dfrac14=\dfrac18&\text{for }w\in\{0,1,2,3,4,5,6,7\}\\\\0&\text{otherwise}\end{cases}

b. The histogram is nothing special, same as the discrete uniform distribution over the interval [0, 7].

3 0
4 years ago
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