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eimsori [14]
3 years ago
7

1 A snail travels of an inch in an hour. How long will it take to travel 5 inches?

Mathematics
2 answers:
kondaur [170]3 years ago
5 0

Answer:

5 hours

Step-by-step explanation:

1in per hour ×5=5 hours

ladessa [460]3 years ago
5 0

Answer: 5 inches

Step-by-step explanation:

If the snail travels an inch {1} an hour the five inches should be 5 hours

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I’m horrible at math and I need to study for the TSI. Help? 4/5+1/10+5/6
Lesechka [4]

Answer:

4 ÷ 5 + 1 ÷ 10 + 5 ÷ 6 = 1.733

Step-by-step explanation:

3 0
2 years ago
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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}}

P(second die (green) | two 1's and two 4's )  = \dfrac{0.5 \times 0.04166666667}{0.02314814815}

P(second die (green) | two 1's and two 4's )  = 0.9

Thus; the probability the die chosen was green is 0.9

8 0
3 years ago
Can someone please help! i have no clue what im doing
gulaghasi [49]

g(x) = -3x - 8


g(x) = 10

⇒ -3x - 8 = 10

⇒ -3x = 18

⇒ x= -6


g(-6) = 10 <==== answer is -6

7 0
3 years ago
The admission fee to an amusement park is $38. It costs an additional p dollars to rent a locker to hold your belongings. The to
Anni [7]
38(4)+p(4)=210

152+p(4)=210

P(4)=58


58/4=14.5


It cost $14.5 for one person to rent a locker
4 0
3 years ago
A desk drawer contains the different-colored markers listed below.
Maksim231197 [3]

Answer:

<h2>total no of markers :36</h2>

in first condition

<h2>no of blue marksers :4</h2>

so probability:

\frac{4}{36}  =  \frac{1}{9}

same goes to the second condition

<h2><em><u>hope</u></em><em><u> it</u></em><em><u> helps</u></em><em><u> you</u></em><em><u><</u></em><em><u>3</u></em></h2>
4 0
2 years ago
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