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AfilCa [17]
3 years ago
6

A fair coin is flipped 10 times and lands on heads 8 times. Provide a reason to justify the difference between the experimental

and theoretical
probabilities. Use the drop-down menus to explain your answer.
There should be a
Choose...
number of trials. With Choose...
flips of the coin, the experimental probability will likely
approach the theoretical probability of Choose...
​
Mathematics
1 answer:
professor190 [17]3 years ago
5 0

Answer:

THere will be 8 heads and 2 tails

Step-by-step explanation:

I don't know

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Need more help guys :(
ololo11 [35]
Last one is correct
good luckkkkkkk
8 0
3 years ago
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A group of 5 students took 15 days to finish a project. After 9 days, 3 students left the group. How many days did it take the s
sladkih [1.3K]

Using proportions, it is found that it took 15 days for the students to finish the project.

<h3>What is a proportion?</h3>

A proportion is a fraction of a total amount, and the measures are related using a rule of three. Due to this, relations between variables, either direct or inverse proportional, can be built to find the desired measures in the problem.

In this problem, we have that:

  • 100% of the project is done by 5 students in 15 days.
  • After the 3 students left, 2 students will have a time of x days to do 100 - (9/15) = 40% of the project.

The compound <em>rule of three</em> is given as follows:

1 project - 5 students - 15 days

0.4 project - 2 students - x days

Increasing the number of students, the number of days is reduced, hence the measures are inverse proportional and the <em>rule of three</em> is given by:

\frac{15}{x} = \frac{2}[5} \times \frac{1}{0.4}

\frac{15}{x} = \frac{2}{2}

15/x = 1

x = 15 days.

It took 15 days for the students to finish the project.

More can be learned about proportions at brainly.com/question/24372153

#SPJ1

6 0
2 years ago
A club puts its members into 5 groups for an activity. After 7 students have to leave early, there are only 3 students left to f
soldier1979 [14.2K]

Answer:

I think that there were 12 students in each group fam.

Step-by-step explanation:

I dont really know so dont get mad at me if i am wrong dude

6 0
3 years ago
Consider a rabbit population​ P(t) satisfying the logistic equation StartFraction dP Over dt EndFraction equals aP minus bP squa
maria [59]

Solution:

Given :

$\frac{dP}{dt}= aP-bP^2$         .............(1)

where, B = aP = birth rate

            D = $bP^2$  =  death rate

Now initial population at t = 0, we have

$P_0$ = 220 ,  $B_0$ = 9 ,  $D_0$ = 15

Now equation (1) can be written as :

$ \frac{dP}{dt}=P(a-bP)$

$\frac{dP}{dt}=bP(\frac{a}{b}-P)$    .................(2)

Now this equation is similar to the logistic differential equation which is ,

$\frac{dP}{dt}=kP(M-P)$

where M = limiting population / carrying capacity

This gives us M = a/b

Now we can find the value of a and b at t=0 and substitute for M

$a_0=\frac{B_0}{P_0}$    and     $b_0=\frac{D_0}{P_0^2}$

So, $M=\frac{B_0P_0}{D_0}$

          = $\frac{9 \times 220}{15}$

          = 132

Now from equation (2), we get the constants

k = b = $\frac{D_0}{P_0^2} = \frac{15}{220^2}$

        = $\frac{3}{9680}$

The population P(t) from logistic equation is calculated by :

$P(t)= \frac{MP_0}{P_0+(M-P_0)e^{-kMt}}$

$P(t)= \frac{132 \times 220}{220+(132-220)e^{-\frac{3}{9680} \times132t}}$

$P(t)= \frac{29040}{220-88e^{-\frac{396}{9680} t}}$

As per question, P(t) = 110% of M

$\frac{110}{100} \times 132= \frac{29040}{220-88e^{\frac{-396}{9680} t}}$

$ 220-88e^{\frac{-99}{2420} t}=200$

$ e^{\frac{-99}{2420} t}=\frac{5}{22}$

Now taking natural logs on both the sides we get

t = 36.216

Number of months = 36.216

8 0
4 years ago
I think it’s A am I right
Mandarinka [93]

Pi(r)(r)(2) + pi(diameter)(height)

3.14(4)(4)(2)+ 3.14(8)(1.5)

138.16

4 0
3 years ago
Read 2 more answers
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