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kobusy [5.1K]
3 years ago
8

Consider the polynomial how many terms are there​

Mathematics
1 answer:
garri49 [273]3 years ago
6 0

Answer: 6 terms

Step-by-step explanation:

The terms are 7kn, 2k^2, -2k, -2n, -5, and 5n^2.

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Please help if you can ik it is Pythagorean theorem
kicyunya [14]

Step-by-step explanation:

A = (1/2)bh ---> h = 2A/b = 2(12 cm^2)/(5 cm) = 4.80 cm

---> x^2 = h^2 + (b/2)^2

= (4.8 cm)^2 + (2.5)^2

= 23.04 cm^2 + 6.25 cm^2

or x = 5.41 cm

Therefore, the perimeter P is

P = 2x + b = 2(5.41 cm) + 5 cm = 15.8 cm

5 0
3 years 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}}

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
Pls answer fast, i need to go to friends but parents wont let me.
Tanya [424]

Answer:

line C

Step-by-step explanation:

The equation would be

y= 1/2x

So that means the slope is 1/2

Starting at (0,0) we would go up 1 and over 2

That would be point (1,2)

That is the purple line, line C

4 0
3 years ago
If the diagonal of a square is 11.3 meters, approximidley<br> what is the perimeter of the square?
Andrei [34K]

Answer:

About 9.6 meters

Step-by-step explanation:

Refer to attachment

8 0
3 years ago
A music school charges a registration fee in addition to a fee per lesson. Write an equation if each lesson costs $40 and a regi
Jobisdone [24]
T = total cost for music school
L = # of lessons

T = $1,190 + ($40 × L)
T = $1,190 + ($40 × 12)
T. = $1,190 + $480
T = $ 1,670 for 12 lessons
3 0
3 years ago
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