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3 years ago

What does 49.39 round to.

1 answer:

4 0

If you want to round 49.39 to the nearest tenths place you will have to look at your tenths digit. The digit is 3, right? Look at the next number after 3. It is the number 9 for the hundredths place. If the number is bigger than 5 you round it. If it is smaller, you keep it the same. 9 is bigger than 5 so you make the 3 into a 4. Then, you're answer is 49.4 or 49.40 (Which are basically the same answer) It also works the same with the other digits. Simple, right? :)

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Suppose you pay a dollar to roll two dice. if you roll 5 or a 6 you Get your dollar back +2 more just like it the goal will be t

LiRa [457]

Answer:

(a)$67

(b)You are expected to win 56 Times

(c)You are expected to lose 44 Times

Step-by-step explanation:

The sample space for the event of rolling two dice is presented below

$(1,1), (2,1), (3,1), (4,1), (5,1), (6,1)\\(1,2), (2,2), (3,2), (4,2), (5,2), (6,2)\\(1,3), (2,3), (3,3), (4,3), (5,3), (6,3)\\(1,4), (2,4), (3,4), (4,4), (5,4), (6,4)\\(1,5), (2,5), (3,5), (4,5), (5,5), (6,5)\\(1,6), (2,6), (3,6), (4,6), (5,6), (6,6)$

Total number of outcomes =36

The event of rolling a 5 or a 6 are:

$(5,1), (6,1)\\ (5,2), (6,2)\\( (5,3), (6,3)\\ (5,4), (6,4)\\(1,5), (2,5), (3,5), (4,5), (5,5), (6,5)\\(1,6), (2,6), (3,6), (4,6), (5,6), (6,6)$

Number of outcomes =20

Therefore:

P(rolling a 5 or a 6) $=\dfrac{20}{36}$

The probability distribution of this event is given as follows.

$\left|\begin{array}{c|c|c}$Amount Won(x)&-\$1&\$2\\&\\P(x)&\dfrac{16}{36}&\dfrac{20}{36}\end{array}\right|$

First, we determine the expected Value of this event.

Expected Value

$=(-\$1\times \frac{16}{36})+ (\$2\times \frac{20}{36})\\=\$0.67$

Therefore, if the game is played 100 times,

Expected Profit =$0.67 X 100 =$67

If you play the game 100 times, you can expect to win $67.

(b)

Probability of Winning $=\dfrac{20}{36}$

If the game is played 100 times

Number of times expected to win

$=\dfrac{20}{36} \times 100\\=56$ times$

Therefore, number of times expected to loose

= 100-56

=44 times

8 0

3 years ago

Which situation could be represented by the equation 32.4 = 8.1x?

Oduvanchick [21]

Answer:

C or A

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

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ad-work [718]

10 and 15 dear^^ 5(15) + 3(15) =28

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2 years ago

This table represents the number of ounces, u, a kitten grows after w, weeks. When graphed, all of the points in the table lie o

Ira Lisetskai [31]

Answer:slope =3/4 y-intercept=3

Step-by-step explanation:

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3 years ago

Find the vertical and horizontal asymptotes, domain, range, and roots of f (x) = -1 / x-3 +2.

gladu [14]

Answer:

Vertical asymptote: $x=3$

Horizontal asymptote: $f(x) =2$

Domain of f(x) is all real numbers except 3.

Range of f(x) is all real numbers except 2.

Step-by-step explanation:

Given:

Function:

$f (x) = -\dfrac{1 }{ x-3} +2$

One root, $x = 3.5$

To find:

Vertical and horizontal asymptote, domain, range and roots of f(x).

Solution:

First of all, let us find the roots of f(x).

<em>Roots of f(x) means the value of x where f(x) = 0</em>

$0= -\dfrac{1 }{ x-3} +2\\\Rightarrow 2= \dfrac{1 }{ x-3}\\\Rightarrow 2x-2 \times 3=1\\\Rightarrow 2x=7\\\Rightarrow x = 3.5$

One root, $x = 3.5$

Domain of f(x) i.e. the values that we give as input to the function and there is a value of f(x) defined for it.

For x = 3, the value of f(x) $\rightarrow \infty$

For all, other values of $x$ , $f(x)$ is defined.

Hence, Domain of f(x) is all real numbers except 3.

Range of f(x) i.e. the values that are possible output of the function.

f(x) = 2 is not possible in this case because something is subtracted from 2. That something is $\frac{1}{x-3}$ .

Hence, Range of f(x) is all real numbers except 2.

Vertical Asymptote is the value of x, where value of f(x) $\rightarrow \infty$ .

$-\dfrac{1 }{ x-3} +2 \rightarrow \infty$

It is possible only when

$x-3=0\\\Rightarrow x=3$

$\therefore$ vertical asymptote: $x=3$

Horizontal Asymptote is the value of f(x) , where value of x $\rightarrow \infty$ .

$x\rightarrow \infty \Rightarrow \dfrac{1 }{ x-3} \rightarrow 0\\\therefore f(x) =-0+2 \\\Rightarrow f(x) =2$

$\therefore$ Horizontal asymptote: $f(x) =2$

Please refer to the graph of given function as shown in the attached image.

5 0

3 years ago