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3 years ago

Please answer this question now

Mathematics

1 answer:

fomenos3 years ago

8 0

Answer:

95 degrees

Step-by-step explanation:

Measure of arc AB is 360 - (101+97+69) = 360 - 267 = 93 degrees.

Measure of arc DAB is 97+93 = 190 degrees, so measure of angle C is 190/2 = 95 degrees.

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Which graph would help you find the solutions of 2 – x2 = –x?

Lelechka [254]

Step-by-step explanation:

A quadratic graph will surely help you find the solutions

X2-x-2=0

x1=2 x2= -1

Substitute each of the x value in the original equation to get the y value

(2)^2-(2)-2= 0 (2,0)

(-1)^2-(-1)-2=-2 (-1,-2)

plot the coordinates on the graph,the graph should be a curve

6 0

3 years ago

Find the distance between the points given. (-3, 2) and (9, -3) A √37 B √61 C 13

Kitty [74]

Answer:

$\huge{ \fbox{ \sf{13 \: units}}}$

Option C is correct.

Step-by-step explanation:

$\star{ \sf{ \: Let \: the \: points \: be \: a \: and \: b}}$

$\star{ \sf{ \: let \: A(-3, 2) \: be \: (x1 \:, y1) \: and \: B(9, -3) \: be \: (x2 \:, y2) }}$

$\underline{ \sf{Finding \: the \: distance \: between \: the \: given \: points}} :$

$\boxed{ \sf{Distance = \sqrt{ {(x2 - x1)}^{2} + {(y2 - y1)}^{2} } }}$

$\mapsto{ \sf{ \sqrt{ {(9 - ( - 3))}^{2} + {( - 3 - 2)}^{2} } }}$

$\mapsto{ \sf{ \sqrt{ {(9 + 3)}^{2} + {( - 3 - 2)}^{2} }}}$

$\mapsto{ \sf{ \sqrt{ {(12)}^{2} + {( - 5)}^{2} } }}$

$\mapsto{ \sf{ \sqrt{144 + 25}}}$

$\mapsto{ \sf{ \sqrt{169} }}$

$\mapsto{ \sf{ \sqrt{ {(13)}^{2} } }}$

$\mapsto{ \sf{ 13 \: units}}$

Hope I helped!

Best regards! :D

~ $\text{TheAnimeGirl}$

7 0

3 years ago

suppose you write an equation that gives "a" as a function of "b". Which is the dependent variable and which is the independent

blondinia [14]

If im not wrong which i may not the
Dependent is A
Independent is B

4 0

3 years ago

Read 2 more answers

THE VOLUME OF A SPHERE IS 38808cm^3. find the height of the cone

Mars2501 [29]

I am not 100% positive but I think it is 4117.66

8 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