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Serjik [45]
3 years ago
8

A box has dimensions of 17 inches long.1.3 feet wide, and 8 inches high. What is the valine of the box? The formula for the volu

me is V = l• w •h.
Mathematics
1 answer:
andreev551 [17]3 years ago
6 0

convert the feet to inches so all 3 dimensions are in the same measurement

1.3 x 12 = 15.6 inches

now multiply them all

17 x 15.6 x 8 = 2121.6 cubic inches

round the answer if you need to

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Help !!!!!!!!!!! me part 2 question read 1 then read 2 I andswer one need help with 2
sergiy2304 [10]

Answer:

The volume of a cylinder is given by πr²h where, r is the radius of the cylinder and h is the height of the cylinder. Also r=d/2 , where d is the diameter of the cylinder. Therefore the volume becomes one-fourth of the initial volume. ... The new volume is 730 mL.

7 0
3 years ago
What is the slope of the line that passes through the points (-9, -8)((−15,−16)? Write your answer in simplest form.
alexandr402 [8]

Answer:

Therefore the slope of the line that passes through the points (-9, -8),(−15,−16) is

Slope=\dfrac{4}{3}

Step-by-step explanation:

Given:

Let,

point A( x₁ , y₁) ≡ ( -9 ,-8)

point B( x₂ , y₂ )≡ (-15 ,-16)

To Find:  

Slope = ?

Solution:

Slope of Line Segment AB is given as

Slope(AB)=\dfrac{y_{2}-y_{1} }{x_{2}-x_{1} }

Substituting the values we get

Slope(AB)=\dfrac{-16-(-8)}{-15-(-9)}\\\\Slope(AB)=\dfrac{-16+8}{-15+9}\\\\Slope(AB)=\dfrac{-8}{-6}=\dfrac{4}{3}

Therefore the slope of the line that passes through the points (-9, -8),(−15,−16) is

Slope(AB)=\dfrac{4}{3}

4 0
3 years ago
A taxi driver wrote a formula to represent the mileage and fee for his last customer. He wrote the equation 3.00 + 0.50m =9.50,
Lemur [1.5K]
3.00 + 0.50m = 9.50

2. true 
5. true


4 0
3 years ago
Read 2 more answers
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
What is the average rate of change between f (1) and f (5) in the function f (x) = x2 - x - 6 ?
NISA [10]

Answer:

{ \tt{average =  \frac{f(1) + f(5)}{2} }} \\  \\  = { \tt{ \frac{( {1}^{2}  - 1 - 6) + ( {5}^{2} - 5 - 6) }{2} }} \\  =  \frac{ - 6 + 14}{2}  \\  \\  = { \tt{4}}

5 0
2 years ago
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