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3 years ago

Choose the function that the graph represents

Mathematics

1 answer:

Reil [10]3 years ago

5 0

I honestly don't know if this is correct but it might be i think its the middle one y=f(x)=x to the second power

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Triangle ABC has angle measures as shown. What is the value of X? Show your calculations.

dezoksy [38]

180-96= 84
That means 8x+12=84
84-12=72
That means 8x=72
72÷8=9
therefore X=9

8 0

3 years ago

Which symbol makes the comparison true? 0.74 to 0.740

DENIUS [597]

Answer:

Step-by-step explanation:

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2 years ago

The equation of a circle in general form is x2+y2+22x+14y−55=0 . What is the equation of the circle in standard form?

ra1l [238]

$x^2+y^2+22x+14y-55=0 \\\\x^2+22x+y^2+14y=55\\\\(x^2+22x+121)-121+(y^2+14y+49)-49=55\\\\(x+11)^2+(y+7)^2-121-49=55\\\\
(x+11)^2+(y+7)^2-170=55\\\\(x+11)^2+(y+7)^2=55+170\\\\\boxed{(x+11)^2+(y+7)^2=225}$

Answer B.

4 0

2 years ago

Read 2 more answers

Pleases help quick please

Solnce55 [7]

Answer:

1 and 2

Step-by-step explanation:

$\sqrt{2}$ = 1.41421356237

3 0

3 years ago

Read 2 more answers

The sample space of a random experiment is {a, b, c, d, e} with probabilities 0.1, 0.1, 0.2, 0.4, and 0.2 respectively. Let A de

antoniya [11.8K]

Answer:

a) $P(A) =P(a)+P(b) +P(c)= 0.1+0.1+0.2 = 0.4$

b) $P(B) =P(c) +P(d)+P(e)=0.2+0.4+0.2=0.8$

c) $P(A') = 1-P(A) =1-0.4=0.6$

d) $P(A \cup B) =0.4 +0.8-0.2 =1.0$

e) The intersection between the set A and B is the element c so then we have this:

$P(A \cap B) = P(c) =0.2$

Step-by-step explanation:

We have the following space provided:

$S= [a,b,c,d,e]$

With the following probabilities:

$P(a) =0.1, P(b)=0.1, P(c) =0.2, P(d)=0.4, P(e)=0.2$

And we define the following events:

A= [a,b,c], B=[c,d,e]

For this case we can find the individual probabilities for A and B like this:

$P(A) = 0.1+0.1+0.2 = 0.4$

$P(B) =0.2+0.4+0.2=0.8$

Determine:

a. P(A)

$P(A) =P(a)+P(b) +P(c)= 0.1+0.1+0.2 = 0.4$

b. P(B)

$P(B) =P(c) +P(d)+P(e)=0.2+0.4+0.2=0.8$

c. P(A’)

From definition of complement we have this:

$P(A') = 1-P(A) =1-0.4=0.6$

d. P(AUB)

Using the total law of probability we got:

$P(A \cup B) =P(A) +P(B)-P(A \cap B)$

For this case $P(A \cap B) = P(c) =0.2$ , so if we replace we got:

$P(A \cup B) =0.4 +0.8-0.2 =1.0$

e. P(AnB)

The intersection between the set A and B is the element c so then we have this:

$P(A \cap B) = P(c) =0.2$

8 0

3 years ago