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2 years ago

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Select the system of inequalities that has been graphed below.

1 answer:

Olenka [21]2 years ago

7 0

The answer is c. Hope that helps!!

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In ⊙O, ST and VT are tangents. m∠STV = 22°. Find the value of a, b, and m∠SOV.

Rasek [7]

Answer:

$\huge \orange {\boxed {a =202\degree}}$

$\huge \purple {\boxed { b = 158\degree}}$

$\huge \red {\boxed {m\angle SOV = 158\degree}}$

Step-by-step explanation:

In $\odot$ O, ST and VT are tangents at points S and V respectively.

$\therefore OS\perp ST, \:and\: OV\perp VT$

$\therefore m\angle OST=m\angle OVT = 90\degree$

In quadrilateral OSTV,

$m\angle SOV +m\angle OST+m\angle OVT+m\angle STV = 360\degree$

(By interior angle sum postulate of a quadrilateral)

$m\angle SOV +90\degree +90\degree +22\degree = 360\degree$

$m\angle SOV +202\degree = 360\degree$

$m\angle SOV = 360\degree-202\degree$

$\huge \red {\boxed {m\angle SOV = 158\degree}}$

$\because b = m\angle SOV$

(Measure of minor arc is equal to measure of its corresponding central angle)

$\huge \purple {\boxed {\therefore b = 158\degree}}$

$\because a + b= 360\degree$

(By arc sum property of a circle)

$\therefore a = 360\degree - b$

$\therefore a = 360\degree -158\degree$

$\huge \orange {\boxed {\therefore a =202\degree}}$

6 0

2 years ago

Weary of the low turnout in student elections, a college administration decides to choose an SRS of three students to form an ad

-Dominant- [34]

Answer:

$P(ABC) = 0.110592$

$P(ABC^c) = 0.119808$

$P(AB^cC) = 0.119808$

$P(A^cBC) = 0.119808$

$P(AB^cC^c) = 0.129792$

$P(A^cBC^c) = 0.129792$

$P(A^cB^cC) = 0.129792$

$P(A^cB^cC^c) = 0.140608$

Step-by-step explanation:

Given

$P(A) = P(B) = P(C) = 48\%$

Convert the probability to decimal

$P(A) = P(B) = P(C) = 0.48$

Solving (a): P(ABC)

This is calculated as:

$P(ABC) = P(A) * P(B) * P(C)$

This gives:

$P(ABC) = 0.48*0.48*0.48$

$P(ABC) = 0.110592$

Solving (b): $P(ABC^c)$

This is calculated as:

$P(ABC^c) = P(A) * P(B) * P(C^c)$

In probability:

$P(C^c) = 1 - P(C)$

So, we have:

$P(ABC^c) = P(A) * P(B) * (1 - P(C))$

$P(ABC^c) = 0.48 * 0.48 * (1 - 0.48)$

$P(ABC^c) = 0.48 * 0.48 * 0.52$

$P(ABC^c) = 0.119808$

Solving (c): $P(AB^cC)$

This is calculated as:

$P(AB^cC) = P(A) * P(B^c) * P(C)$

$P(AB^cC) = P(A) * [1 - P(B)] * P(C)$

$P(AB^cC) = 0.48 * (1 - 0.48)* 0.48$

$P(AB^cC) = 0.48 * 0.52* 0.48$

$P(AB^cC) = 0.119808$

Solving (d): $P(A^cBC)$

This is calculated as:

$P(A^cBC) = P(A^c) * P(B) * P(C)$

$P(A^cBC) = [1-P(A)] *P(B) * P(C)$

$P(A^cBC) = (1 - 0.48)* 0.48 * 0.48$

$P(A^cBC) = 0.52* 0.48 * 0.48$

$P(A^cBC) = 0.119808$

Solving (e): $P(AB^cC^c)$

This is calculated as:

$P(AB^cC^c) = P(A) * P(B^c) * P(C^c)$

$P(AB^cC^c) = P(A) * [1-P(B)] * [1-P(C)]$

$P(AB^cC^c) = 0.48 * [1-0.48] * [1-0.48]$

$P(AB^cC^c) = 0.48 * 0.52*0.52$

$P(AB^cC^c) = 0.129792$

Solving (f): $P(A^cBC^c)$

This is calculated as:

$P(A^cBC^c) = P(A^c) * P(B) * P(C^c)$

$P(A^cBC^c) = [1-P(A)] * P(B) * [1-P(C)]$

$P(A^cBC^c) = [1-0.48] * 0.48 * [1-0.48]$

$P(A^cBC^c) = 0.52 * 0.48 * 0.52$

$P(A^cBC^c) = 0.129792$

Solving (g): $P(A^cB^cC)$

This is calculated as:

$P(A^cB^cC) = P(A^c) * P(B^c) * P(C)$

$P(A^cB^cC) = [1-P(A)] * [1-P(B)] * P(C)$

$P(A^cB^cC) = [1-0.48] * [1-0.48] * 0.48$

$P(A^cB^cC) = 0.52 * 0.52 * 0.48$

$P(A^cB^cC) = 0.129792$

Solving (h): $P(A^cB^cC^c)$

This is calculated as:

$P(A^cB^cC^c) = P(A^c) * P(B^c) * P(C^c)$

$P(A^cB^cC^c) = [1-P(A)] * [1-P(B)] * [1-P(C)]$

$P(A^cB^cC^c) = [1-0.48] * [1-0.48] * [1-0.48]$

$P(A^cB^cC^c) = 0.52*0.52*0.52$

$P(A^cB^cC^c) = 0.140608$

5 0

2 years ago

HELP QUICK PLEASE I WILL MARK BRAINLIEST

Minchanka [31]

Answer:

Angle L would still be 45 degrees because the size of the triangle did not change when it was moved

3 0

2 years ago

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Write and then evaluate an expression for the following number line

drek231 [11]

Answer:

ok there are know boxes

Step-by-step explanation:

6 0

2 years ago

How to solve this equation 18.5 = -4.5 + 7(k + 6)

yawa3891 [41]

1. Add -4.5 to 18.5
2. Divide 23 by 7
3. Then subtract 6 from -2.7

8 0

2 years ago

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