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coldgirl [10]
3 years ago
12

A thumbtack that is tossed can land point up or point down. The probability of a tack landing point up is 0.2. A simulation was

conducted in which a trial consisted of tossing 5 thumbtacks and recording the number of thumbtacks that land point up. Many trials of the simulation were conducted and the results are shown in the histogram.
Based on the results of the simulation, which of the following is closest to the probability that at least 2 thumbtacks land pointing up when 5 thumbtacks are tossed?

A 0.09

B 0.19

C 0.28

D 0.72

E 0.91

Mathematics
2 answers:
sweet-ann [11.9K]3 years ago
8 0

Answer:

Option C: 0.28

Step-by-step explanation:

This is a binomial probability distribution problem.

Now, we want to find the probability that at least 2 thumbtacks land pointing up when 5 thumbtacks are tossed. This is written as;

P(X ≥ 2) = P(2) + P(3) + P(4) + P(5)

From the histogram;

P(5) = 0.02

P(4) = 0.02

P(3) = 0.05

P(2) = 0.19

Thus;

P(X ≥ 2) = 0.19 + 0.05 + 0.02 + 0.02

P(X ≥ 2) = 0.28

Delvig [45]3 years ago
7 0

Answer:

0.28

Step-by-step explanation:

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How many solutions does the following equation have? -4-7+10x=-7+6x
stealth61 [152]

Answer:

Step-by-step explanation:

-4 - 7 + 10x = -7 + 6x...combine like terms

10x - 11 = 6x - 7

10x - 6x = -7 + 11

4x = 4

x = 4/4

x = 1.........ONE SOLUTION

4 0
3 years ago
Suppose the weights of the Boxers at this club are Normally distributed with a mean of 166 pounds and a standard deviation of 5.
lesya [120]

Answer:

0.1994 is the required probability.      

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 166 pounds

Standard Deviation, σ = 5.3 pounds

Sample size, n = 20

We are given that the distribution of weights is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

Standard error due to sampling =

=\dfrac{\sigma}{\sqrt{n}} = \dfrac{5.3}{\sqrt{20}} = 1.1851

P(sample of 20 boxers is more than 167 pounds)

P( x > 167) = P( z > \displaystyle\frac{167 - 166}{1.1851}) = P(z > 0.8438)

= 1 - P(z \leq 0.8438)

Calculation the value from standard normal z table, we have,  

P(x > 167) = 1 - 0.8006= 0.1994 = 19.94\%

0.1994 is the probability that the mean weight of a random sample of 20 boxers is more than 167 pounds

3 0
3 years ago
Each line represents a proportional relationship. Write an equation for each line. Show your work.
tiny-mole [99]

Answer:

The equation of line a is y = x

The equation of line b is y =  \frac{2}{3} x

Step-by-step explanation:

The equation of the proportional is y = m x, where

  • m is the slope of the line (constant of proportionality)

The rule of the slope of a line is m = \frac{y2-y1}{x2-x1} , where

  • (x1, y1) and (x2, y2) are two points on the line

∵ Line a passes through points (0, 0) and (3, 3)

∴ x1 = 0 and y1 = 0

∴ x2 = 3 and y2 = 3

→ Substitute them in the rule of the slope above

∵ m = \frac{3-0}{3-0}=\frac{3}{3}

∴ m = 1

→ Substitute in the form of the equation above

∴ y = (1)x

∴ y = x

∴ The equation of line a is y = x

∵ Line b passes through points (0, 0) and (3, 2)

∴ x1 = 0 and y1 = 0

∴ x2 = 3 and y2 = 2

→ Substitute them in the rule of the slope above

∵ m = \frac{2-0}{3-0}=\frac{2}{3}

∴ m = \frac{2}{3}

→ Substitute in the form of the equation above

∴ y = (\frac{2}{3}) x

∴ y =  \frac{2}{3} x

∴ The equation of line b is y =  \frac{2}{3} x

7 0
3 years ago
The daily mean temperature in a particular place is 83. how many cooling-degree days were accumulated?
olga55 [171]
Degree days are the difference between the daily temperature mean, (high temperature plus low temperature divided by two) and 65°F. 65°F because we assume that at this temperature we do not need cooling or heating. 
In our case the daily mean temperature is 83°F and X is the number of cooling-degree days.
X=83-65
X=18
4 0
3 years ago
I need help! Will this would be very appreciated!pls
elena-s [515]

Answer:

No real

solution

Step-by-step explanation:

Firstly, let us check if we would be having a real solution

We start by rewriting the equation

We have this as;

8x^2 -25x + 24 = 0

We proceed to get the discriminant

Mathematically, we have this as;

D = b^2 - 4ac

b is the coefficient of x which is -25

a is the coefficient of x^2 which is 8

c is the last number which is 24

So we have;

D = (-25)^2 - 4(8)(24)

D = 625 - 768 = -143

Since the value of the discriminant is negative, there cannot be real roots

What we have as solution are complex roots

6 0
2 years ago
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