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user100 [1]
2 years ago
11

Fred makes $750 a week working 5 days .How much does he make in one day

Mathematics
1 answer:
Radda [10]2 years ago
6 0

Answer:

$150

Step-by-step explanation:

750/5 = 150

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What is the value of x ​
SVEN [57.7K]

Using the pythagorean theorem, the value of x would be eight. The equation of Pythagorean is

a^2 + b^2 = c^2

(C is the hypotenuse. Hypotenuse being the length of 10) Applying the lengths leaves us with the equation

36 + x^2 = 100.

You then subtract 36 from 100 to get X alone on the opposite side of the equation.

X^2 = 64

Lastly, you take the square root of 64. Therefore, the final answer will be eight.

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Graph the line with the equation <br> y = -1/3x + 4.
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Answer in the image

Step-by-step explanation:

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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

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Answer:

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First find the area of the whole pizza.

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The diameter is given as 131 feet, so r would be 131/2 = 65.5 feet.

I will use 3.14 for PI since the problem doesn't say what value to use.


Area = PI x 65.5^2

Area = 3.14 x 4290.25

Area = 13,471.39 square feet


Divide the total area by the number of pieces:

13,471.39 / 50 = 269.43 square feet


To find the length of the crust first find the circumference of the whole pizza:

Circumference = PI x diameter

Circumference = 3.14 x 131 = 411.34 feet

Now divide the total by the number of pieces:

411.34 / 50 = 8.23 feet


Each slice has an area of about 269.43 square feet and a crust length of 8.23 feet.


8 0
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