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exis [7]
3 years ago
12

find the probability of at least 3 successes in a 3 trials of a binomial experiment i which the probability of successe is 50%

Mathematics
1 answer:
katrin2010 [14]3 years ago
7 0

each trial is independent so do

.5 x .5 x .5 or .5^3

.125 is the probability.

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Pls help question in image​
cluponka [151]
Answer: 41.16 cm cubed.

= (0.5 • 4.2 • 2.8) • 7 = 41. 16

Explanation: B = 1/2 base times hight of the triangle. Not the figure.
Remember the figure is also tipped over, and its not up right. So you’ll have to try to imagine and image it standing up right.
6 0
1 year ago
(5.2x10^-1)(1.5x10^-5)
VLD [36.1K]

Answer:

(10^-6) (5.2x1.5 )=10^-6 x 7.8 = 7.8 x10^-6

Step-by-step explanation:

0.000001 x7.8=0.0000078

6 0
2 years ago
You have to evaluate<br> 3³+11-10÷2×4=w
denis23 [38]

Answer:

18

<u><em>Brainliest, please!</em></u>

Step-by-step explanation:

3^3 + 11 - 10 / 2 x 4 = w

27 + 11 - 10 / 2 x 4 = w

27 + 11 - 5 x 4 = w

27 + 11 - 20 = w

38 - 20 = w

w = 18

3 0
3 years ago
Help Mr plz I'm tell you I think it 8
katrin2010 [14]
It is 9 weeks. 8 times 12 is only 96$
7 0
3 years ago
An area is approximated to be 14 in 2 using a left-endpoint rectangle approximation method. A right- endpoint approximation of t
USPshnik [31]
The trapezoidal approximation will be the average of the left- and right-endpoint approximations.

Let's consider a simple example of estimating the value of a general definite integral,

\displaystyle\int_a^bf(x)\,\mathrm dx

Split up the interval [a,b] into n equal subintervals,

[x_0,x_1]\cup[x_1,x_2]\cup\cdots\cup[x_{n-2},x_{n-1}]\cup[x_{n-1},x_n]

where a=x_0 and b=x_n. Each subinterval has measure (width) \dfrac{a-b}n.

Now denote the left- and right-endpoint approximations by L and R, respectively. The left-endpoint approximation consists of rectangles whose heights are determined by the left-endpoints of each subinterval. These are \{x_0,x_1,\cdots,x_{n-1}\}. Meanwhile, the right-endpoint approximation involves rectangles with heights determined by the right endpoints, \{x_1,x_2,\cdots,x_n\}.

So, you have

L=\dfrac{b-a}n\left(f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1})\right)
R=\dfrac{b-a}n\left(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n)\right)

Now let T denote the trapezoidal approximation. The area of each trapezoidal subdivision is given by the product of each subinterval's width and the average of the heights given by the endpoints of each subinterval. That is,

T=\dfrac{b-a}n\left(\dfrac{f(x_0)+f(x_1)}2+\dfrac{f(x_1)+f(x_2)}2+\cdots+\dfrac{f(x_{n-2})+f(x_{n-1})}2+\dfrac{f(x_{n-1})+f(x_n)}2\right)

Factoring out \dfrac12 and regrouping the terms, you have

T=\dfrac{b-a}{2n}\left((f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1}))+(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n))\right)

which is equivalent to

T=\dfrac12\left(L+R)

and is the average of L and R.

So the trapezoidal approximation for your problem should be \dfrac{14+21}2=\dfrac{35}2=17.5\text{ in}^2
4 0
3 years ago
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