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aivan3 [116]
3 years ago
14

Helpppppppp IM TAKING A TESTTTTTT

Mathematics
1 answer:
vampirchik [111]3 years ago
6 0

Answer:

B

Step-by-step explanation:

55 kids like hotdogs divided by 200 total kids = 27.5%

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If (x)= 2x square minus 1,then f(-3)
viva [34]

Answer:

17.

Step-by-step explanation:

f(x) = 2x^2 - 1

f(-3) = 2(-3)^2 - 1

= 2 * 9 - 1

= 18 - 1

= 17

Hope this helps!

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
Sam has ⅙ of a book left to read. He wants to finish the book by reading the same amount every day for 4 days. How much of the b
babymother [125]

4 (1/16)  =  4/16   =  1/4

(1/4) / 4    =   1/16  each day

7 0
3 years ago
Read 2 more answers
What is the slope of the line shown below?<br> у<br> N<br> 2<br> 1-5-4-3-2-1<br> X<br> 1 2 3 4 5
TEA [102]

Answer:2

2/1=2

I hope this is good enough:

4 0
3 years ago
17. Which of the following is a graph of y= x^2?
Juli2301 [7.4K]

Answer:

The parabola - quadratic.

Step-by-step explanation:

There is only graph / option in the picture but it is the correct one.  Regardless, whether x is a positive or negative, the y coordinate is going to be a positive number as x^2 will produce a positive.  So as -x^2 will also be positive.

7 0
3 years ago
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