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lorasvet [3.4K]
3 years ago
11

In 150 spins, about how often do you expect to less than 6?

Mathematics
1 answer:
netineya [11]3 years ago
7 0
It depends on how many numbers there are.
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PLEASE HELP DUE IN 3 minutes
son4ous [18]

Answer:

2....................

3 0
3 years ago
Read 2 more answers
Solve the system of equations: 3x + 6y=6 and 2y - 3x = 6.
weeeeeb [17]

Answer:

x = -1, y = 1.5

Step-by-step explanation:

3x + 6y = 6 -----> x + 2y = 2

-3x + 2y = 6 -----> -3x + 2y = 6

------------------

4x = -4

x = -1, y = 1.5

7 0
1 year ago
The number of phone calls that Actuary Ben receives each day has a Poisson distribution with mean 0.1 during each weekday and me
Dovator [93]

Answer:

There is a 0.73% probability that Ben receives a total of 2 phone calls in a week.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given time interval.

The problem states that:

The number of phone calls that Actuary Ben receives each day has a Poisson distribution with mean 0.1 during each weekday and mean 0.2 each day during the weekend.

To find the mean during the time interval, we have to find the weighed mean of calls he receives per day.

There are 5 weekdays, with a mean of 0.1 calls per day.

The weekend is 2 days long, with a mean of 0.2 calls per day.

So:

\mu = \frac{5(0.1) + 2(0.2)}{7} = 0.1286

If today is Monday, what is the probability that Ben receives a total of 2 phone calls in a week?

This is P(X = 2). So:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 2) = \frac{e^{-0.1286}*0.1286^{2}}{(2)!} = 0.0073

There is a 0.73% probability that Ben receives a total of 2 phone calls in a week.

3 0
3 years ago
=
ivann1987 [24]

9514 1404 393

Answer:

  • large: 55 lb
  • small: 30 lb

Step-by-step explanation:

Let x and y represent the weights of the large and small boxes, respectively. The problem statement gives rise to the system of equations ...

  x + y = 85 . . . . . combined weight of a large and small box

  70x +50y = 5350 . . . . combined weight of 70 large and 50 small boxes

We can subtract 50 times the first equation from the second to find the weight of a large box.

  (70x +50y) -50(x +y) = (5350) -50(85)

  20x = 1100 . . . . simplify

  x = 55 . . . . . . . divide by 20

Using this in the first equation, we can find the weight of a small box.

  55 +y = 85

  y = 30 . . . . . . . subtract 55

A large box weighs 55 pounds; a small box weighs 30 pounds.

4 0
3 years ago
What is the volume of the sphere that has a diameter of 3? use 3.14 for pie​
aev [14]

Hey ! there

Answer:

  • <u>1</u><u>1</u><u>3</u><u>.</u><u>0</u><u>4</u><u> </u><u>unit </u><u>cube</u>

Step-by-step explanation:

In this question we are provided with a sphere <u>having</u><u> </u><u>radius </u><u>3 </u><u>units </u>and <u>value </u><u>of </u><u>π </u><u>is </u><u>3.</u><u>1</u><u>4</u><u> </u><u>.</u><u> </u>And we're asked to find the<u> </u><u>volume</u><u> of</u><u> </u><u>sphere</u><u> </u><u>.</u>

For finding volume of sphere , we need to know its formula . So ,

\qquad \qquad \: \underline{\boxed{ \frak{Volume_{(Sphere)} =  \dfrac{4}{3} \pi r {}^{3} }}}

<u>Where</u><u> </u><u>,</u>

  • π refers to <u>3.</u><u>1</u><u>4</u>

  • r refers to <u>radius</u><u> of</u><u> sphere</u>

<u>Sol</u><u>u</u><u>tion </u><u>:</u><u> </u><u>-</u>

Now , we are substituting value of π and radius in the formula ,

\quad \longrightarrow \qquad \: \dfrac{4}{3}   \times 3.14 \times (3) {}^{3}

Simplifying it ,

\quad \longrightarrow \qquad \: \dfrac{4}{3}  \times 3.14 \times 3 \times 3 \times 3

Cancelling 3 with 3 :

\quad \longrightarrow \qquad \: \dfrac{4}{ \cancel{3}}  \times 3.14 \times 3 \times 3 \times  \cancel{3}

We get ,

\quad \longrightarrow \qquad \:4 \times 3.14 \times 9

Multiplying 4 and 3.14 :

\quad \longrightarrow \qquad \:12.56 \times 9

Multiplying 12.56 and 9 :

\quad \longrightarrow \qquad \:    \pink{\underline{\boxed{\frak{113.04  \: unit \: cube}}}} \quad \bigstar

  • <u>Henceforth</u><u> </u><u>,</u><u> </u><u>volume</u><u> </u><u>of</u><u> </u><u>sphere</u><u> </u><u>having </u><u>radius </u><u>3 </u><u>units </u><u>is </u><em><u>1</u></em><em><u>1</u></em><em><u>3</u></em><em><u> </u></em><em><u>.</u></em><em><u>0</u></em><em><u>4</u></em><em><u> </u></em><em><u>units </u></em><em><u>cube </u></em><em><u>.</u></em>

<h2><u>#</u><u>K</u><u>e</u><u>e</u><u>p</u><u> </u><u>Learning</u></h2>

5 0
2 years ago
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