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Temka [501]
4 years ago
7

A fair coin is tossed three three times in succession. the set of equally likely outcomes is startset hhh comma hht comma hth co

mma thh comma htt comma tht comma tth comma ttt endset{hhh, hht, hth, thh, htt, tht, tth, ttt}. find the probability of getting exactly one tail.
Mathematics
1 answer:
Slav-nsk [51]4 years ago
6 0

Solution: We are given that a fair coin is tossed three times. The sample space associated with the three tosses of fair coin is:

S=\left \{ HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\right \}

We have to find the probability of getting exactly one tail.

From the above sample space, we clearly see there are three outcomes which  favors the probability of exactly one tail.

n( 1 tail) = \left \{HHT,HTHTHH \right \}

Therefore the probability of exactly one tail is

\frac{3}{8} =0.375


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What is the greatest common factor (GCF) of 24x^4 and 80x^5<br> A.) 4x^5<br> B.) 6x^4<br> C.) 8x^4
Volgvan
The greatest common factor (GCF) of the expressions is a product of the GCF of the numerical coefficients and the variables. The GCF of the 24 and 80 is 8. Additionally, the GCF of x^4 and x^5 if x^4. Thus, the GCF is 8x^4. The answer to this problem is letter C. 
3 0
3 years ago
In△XYZ , XY¯¯¯¯¯¯=7in , YZ¯¯¯¯¯=5in , and XZ¯¯¯¯¯=4in . This triangle is reflected across the x-axis to result in △X'Y'Z' . Whic
ki77a [65]

Answer:

B

Step-by-step explanation:

Reflection is a transformation that preserves lengths.

Triangle XYZ, with side lengths XY = 7 in, YZ = 5 in, XZ = 4 in, is reflected across the x-axis to result in △X'Y'Z'.

This means that corresponding sides have the same lengths:

  • XY = X'Y' = 7 in;
  • XZ = X'Z' = 4 in;
  • YZ = Y'Z' = 5 in.

Find the perimeters of both triangles:

P_{XYZ}=XY+XZ+YZ=7+4+5=16\ in\\ \\P_{X'Y'Z'}=X'Y'+X'Z'+Y'Z'=7+4+5=16\ in

Hence, only option B is true.

8 0
3 years ago
Round to the nearest thousand then subtract
dsp73
100

6738 —> 7000
5903 —> 6000

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4 years ago
The amount of a radioactive material changes with time. the table below shows the amount of radioactive material f(t) left after
n200080 [17]
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3 0
3 years ago
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Given that sequence 1.77, 1.92,2.07 is arithmetic sequence.

Using that we have to find 31st term.

To find that we can use nth term formula of arithmetic sequence which is given by:

t_n=a_1+(n-1)d

where n is the number of term so n=31

t_n indicates nth term

a_1 is the first term so a_1=1.77

r is the common difference which is given by difference of consecutive terms

then r=1.92-1.77=0.15

Now plug these values into above formula


t_n=a_1+(n-1)d

t_{31}=1.77+(31-1)*0.15

t_{31}=1.77+(30)*0.15

t_{31}=1.77+4.5

t_{31}=6.27


Hence final answer is 6.27.

8 0
4 years ago
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