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3 years ago

Tess rolls 2 fair dice. What is the probability of obtaining two 4's?

Mathematics

2 answers:

harina [27]3 years ago

4 0

Answer:

1/36

Step-by-step explanation:

Since Tess rolled 2 different dices, there are 36 different possibilities. Rolling two 4's is 1 of them.

Therefore, the probability of Tess rolling 2 fair dice is 1/36

Hope this helped have a great rest of your day!

anygoal [31]3 years ago

3 0

Answer:

1 out of 36 chance because add what a dice equals to then divide

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The length of time taken on the SAT for a group of students is normally distributed with a mean of 2.5 hours and a standard devi

son4ous [18]

Answer:

Step-by-step explanation:

Since the length of time taken on the SAT for a group of students is normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - u)/s

Where

x = length of time

u = mean time

s = standard deviation

From the information given,

u = 2.5 hours

s = 0.25 hours

We want to find the probability that the sample mean is between two hours and three hours.. It is expressed as

P(2 lesser than or equal to x lesser than or equal to 3)

For x = 2,

z = (2 - 2.5)/0.25 = - 2

Looking at the normal distribution table, the probability corresponding to the z score is 0.02275

For x = 3,

z = (3 - 2.5)/0.25 = 2

Looking at the normal distribution table, the probability corresponding to the z score is 0.97725

P(2 lesser than or equal to x lesser than or equal to 3)

= 0.97725 - 0.02275 = 0.9545

6 0

4 years ago

Someone please help me

Phoenix [80]

Answer: 120, 60, 180

Step-by-step explanation:

1) Area of a rectangle's formula is:

l × w

So, looking at the diagram, we get:

6 × 20 = 120 cm²

2) Area of a trapezoid's formula is:

$\frac{a+b}{2}$ × h (where both a and b are the different base lengths)

So, looking at the diagram, we get:

$\frac{10+20}{2}$ × 4 = 60 cm²

3) Add both to find the area of the entire figure.

120 + 60 = 180 cm²

8 0

3 years ago

Find the value of x such that 365 based seven + 43 based x = 217 based 10.

Pepsi [2]

We need to find the base x in the following equation:

$365_7+43_x=217_{10}$

First, lets convert 365 from base 7 to base 10. This is given by

$365_7=3\times7^2+6\times7^1+5\times7^0$

where the upperindex denotes the position of eah number. This gives

$\begin{gathered} 365_7=3\times49+6\times7+5\times1 \\ 365_7=147+42+5 \\ 365_7=194_{10} \end{gathered}$

that is, 365 based 7 is equal to 194 bases 10.

Now, lets do the same for 43 based x. Lets convert 43 based x to base 10:

$43_x=4\times x^1+3\times x^0$

where again, the superindex 0 and 1 denote the position 0 and 1 in the number 43. This gives

$43_x=(4x+3)_{10}$

Now, we have all number in base 10. Then, our first equation can be written in base 10 as

$194_{10}+(4x+3)_{10}=217_{10}$

For simplicity, we can omit the 10 and get

$194+4x+3=217$

so, we can solve this equation for x. By combining similar terms. we have

$197+4x=217$

and by moving 197 to the right hand side, we obtain

$\begin{gathered} 4x=217-197 \\ 4x=20 \end{gathered}$

Finally, we get

$\begin{gathered} x=\frac{20}{4} \\ x=5 \end{gathered}$

Therefore, the solution is x=5

8 0

1 year ago

Someone help meeee??? Pleaseeee

s2008m [1.1K]

Answer:37

Step-by-step explanation:

(5x7)+2

35+2

=37

6 0

3 years ago

Let X denote the amount of time a book on two-hour reserve is actually checked out, and suppose the cdf is the following. F(x) =

Troyanec [42]

Answer:

a) P (x <= 3 ) = 0.36

b) P ( 2.5 <= x <= 3 ) = 0.11

c) P (x > 3.5 ) = 1 - 0.49 = 0.51

d) x = 3.5355

e) f(x) = x / 12.5

f) E(X) = 3.3333

g) Var (X) = 13.8891 , s.d (X) = 3.7268

h) E[h(X)] = 2500

Step-by-step explanation:

Given:

The cdf is as follows:

F(x) = 0 x < 0

F(x) = (x^2 / 25) 0 < x < 5

F(x) = 1 x > 5

Find:

(a) Calculate P(X ≤ 3).

(b) Calculate P(2.5 ≤ X ≤ 3).

(d) What is the median checkout duration ? [solve 0.5 = F()].

(e) Obtain the density function f(x). f(x) = F '(x) =

(f) Calculate E(X).

(g) Calculate V(X) and σx. V(X) = σx =

(h) If the borrower is charged an amount h(X) = X2 when checkout duration is X, compute the expected charge E[h(X)].

Solution:

a) Evaluate the cdf given with the limits 0 < x < 3.

So, P (x <= 3 ) = (x^2 / 25) | 0 to 3

P (x <= 3 ) = (3^2 / 25) - 0

P (x <= 3 ) = 0.36

b) Evaluate the cdf given with the limits 2.5 < x < 3.

So, P ( 2.5 <= x <= 3 ) = (x^2 / 25) | 2.5 to 3

P ( 2.5 <= x <= 3 ) = (3^2 / 25) - (2.5^2 / 25)

P ( 2.5 <= x <= 3 ) = 0.36 - 0.25 = 0.11

c) Evaluate the cdf given with the limits x > 3.5

So, P (x > 3.5 ) = 1 - P (x <= 3.5 )

P (x > 3.5 ) = 1 - (3.5^2 / 25) - 0

P (x > 3.5 ) = 1 - 0.49 = 0.51

d) The median checkout for the duration that is 50% of the probability:

So, P( x < a ) = 0.5

(x^2 / 25) = 0.5

x^2 = 12.5

x = 3.5355

e) The probability density function can be evaluated by taking the derivative of the cdf as follows:

pdf f(x) = d(F(x)) / dx = x / 12.5

f) The expected value of X can be evaluated by the following formula from limits - ∞ to +∞:

E(X) = integral ( x . f(x)).dx limits: - ∞ to +∞

E(X) = integral ( x^2 / 12.5)

E(X) = x^3 / 37.5 limits: 0 to 5

E(X) = 5^3 / 37.5 = 3.3333

g) The variance of X can be evaluated by the following formula from limits - ∞ to +∞:

Var(X) = integral ( x^2 . f(x)).dx - (E(X))^2 limits: - ∞ to +∞

Var(X) = integral ( x^3 / 12.5).dx - (E(X))^2

Var(X) = x^4 / 50 | - (3.3333)^2 limits: 0 to 5

Var(X) = 5^4 / 50 - (3.3333)^2 = 13.8891

s.d(X) = sqrt (Var(X)) = sqrt (13.8891) = 3.7268

h) Find the expected charge E[h(X)] , where h(X) is given by:

h(x) = (f(x))^2 = x^2 / 156.25

The expected value of h(X) can be evaluated by the following formula from limits - ∞ to +∞:

E(h(X))) = integral ( x . h(x) ).dx limits: - ∞ to +∞

E(h(X))) = integral ( x^3 / 156.25)

E(h(X))) = x^4 / 156.25 limits: 0 to 25

E(h(X))) = 25^4 / 156.25 = 2500

8 0

3 years ago