All categories

2 years ago

What is the value of the expression |a + b| + |c| when a = –3, b = 7, and c = –15?

1 answer:

3 0

We know:
$|a|= \left\{\begin{array}{ccc}a&if\ a \ \textgreater \ 0\\0&if\ a=0\\-a&if\ a \ \textless \ 0\end{array}\right\\\\|2|=2\\|0|=0\\|-3|=3$

$|a+b|+|c|$
substitute a = -3; b = 7; c = -15
$|-3+7|+|-15|=|4|+15=4+15=19$

You might be interested in

If the volume of a cube is 64 in3, how long is each side?

8_murik_8 [283]

Answer:

Step-by-step explanation:

4*4*4=16*4=64

Please mark as Brainliest! :)

Have a nice day.

5 0

3 years ago

The half life of radium is 1690 years. if 80 grams are present now, how much will be present in 830 years

Pachacha [2.7K]

Answer:

A(t) = amount remaining in t years

= A0ekt, where A0 is the initial amount and k is a constant to be determined.

Since A(1690) = (1/2)A0 and A0 = 80,

we have 40 = 80e1690k

1/2 = e1690k

ln(1/2) = 1690k

k = -0.0004

So, A(t) = 80e-0.0004t

Therefore, A(430) = 80e-0.0004(430)

= 80e-0.172

≈ 67.4 g

Step-by-step explanation:

5 0

1 year ago

A bag contains two six-sided dice: one red, one green. The red die has faces numbered 1, 2, 3, 4, 5, and 6. The green die has fa

gayaneshka [121]

Answer:

the probability the die chosen was green is 0.9

Step-by-step explanation:

Given that:

A bag contains two six-sided dice: one red, one green.

The red die has faces numbered 1, 2, 3, 4, 5, and 6.

The green die has faces numbered 1, 2, 3, 4, 4, and 4.

From above, the probability of obtaining 4 in a single throw of a fair die is:

P (4 | red dice) = $\dfrac{1}{6}$

P (4 | green dice) = $\dfrac{3}{6}$ = $\dfrac{1}{2}$

A die is selected at random and rolled four times.

As the die is selected randomly; the probability of the first die must be equal to the probability of the second die = $\dfrac{1}{2}$

The probability of two 1's and two 4's in the first dice can be calculated as:

= $\begin {pmatrix} \left \begin{array}{c}4\\2\\ \end{array} \right \end {pmatrix} \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^4$

= $\dfrac{4!}{2!(4-2)!} ( \dfrac{1}{6})^4$

= $\dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^4$

= $6 \times ( \dfrac{1}{6})^4$

= $(\dfrac{1}{6})^3$

= $\dfrac{1}{216}$

The probability of two 1's and two 4's in the second dice can be calculated as:

= $\begin {pmatrix} \left \begin{array}{c}4\\2\\ \end{array} \right \end {pmatrix} \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^2 \times \begin {pmatrix} \dfrac{3}{6} \end {pmatrix} ^2$

= $\dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^2 \times ( \dfrac{3}{6})^2$

= $6 \times ( \dfrac{1}{6})^2 \times ( \dfrac{3}{6})^2$

= $( \dfrac{1}{6}) \times ( \dfrac{3}{6})^2$

= $\dfrac{9}{216}$

∴

The probability of two 1's and two 4's in both dies = P( two 1s and two 4s | first dice ) P( first dice ) + P( two 1s and two 4s | second dice ) P( second dice )

The probability of two 1's and two 4's in both die = $\dfrac{1}{216} \times \dfrac{1}{2} + \dfrac{9}{216} \times \dfrac{1}{2}$

The probability of two 1's and two 4's in both die = $\dfrac{1}{432} + \dfrac{1}{48}$

The probability of two 1's and two 4's in both die = $\dfrac{5}{216}$

By applying Bayes Theorem; the probability that the die was green can be calculated as:

P(second die (green) | two 1's and two 4's ) = The probability of two 1's and two 4's | second dice)P (second die) ÷ P(two 1's and two 4's in both die)

P(second die (green) | two 1's and two 4's ) = $\dfrac{\dfrac{1}{2} \times \dfrac{9}{216}}{\dfrac{5}{216}}$