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3 years ago

Find the area of a sector with a central angle of 120° and a diameter of 7.3 cm. Round to the nearest tenth.

Mathematics

1 answer:

Andru [333]3 years ago

6 0

Answer: $14cm^{2}$

Step-by-step explanation:

diameter=7.3cm

radius= $\frac{diameter}{2}$

= $\frac{7.3}{2}$

=3.65cm

angle =120°

area= $\frac{angle (tita)}{360}$ × $\frac{\pi r^{2} }{1}$

= $\frac{120}{360}$ × $\frac{22}{7}$ × $3.65^{2}$

= $13.95cm^{2}$

≅ $14cm^{2}$

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Erica had $20 she spent on three gifts she spent 9 3/4 dollars on gift a and 4 1/2 dollars and gift b how much money did she hav

77julia77 [94]

Answer:

$5.75

Step-by-step explanation:

20=9.75 + 4.5 + c

20-9.75-4.5=c

5.75=c

6 0

3 years ago

Find the slope and the y-intercept of the line.<br><br> Y=-3/2x+7

goldenfox [79]

The slope is
3/2

Hope this helps

3 0

3 years ago

Read 2 more answers

Seven balls are randomly withdrawn from an urn that contains 12 red, 16 blue, and 18 green balls. Find the probability that (a)

UNO [17]

Answer:

a) P=0.226

b) P=0.6

c) P=0.0008

d) P=0.74

Step-by-step explanation:

We know that the seven balls are randomly withdrawn from an urn that contains 12 red, 16 blue, and 18 green balls. Therefore, we have 46 balls.

a) We calculate the probability that are 3 red, 2 blue, and 2 green balls.

We calculate the number of possible combinations:

$C_7^{46}=\frac{46!}{7!(46-7)!}=53524680$

We calculate the number of favorable combinations:

$C_3^{12}\cdot C_2^{16}\cdot C_2^{18}=660\cdot 120\cdot 153=12117600$

Therefore, the probability is

$P=\frac{12117600}{53524680}\\\\P=0.226$

b) We calculate the probability that are at least 2 red balls.

We calculate the probability withdrawn of 1 or none of the red balls.

We calculate the number of possible combinations:

$C_7^{46}=\frac{46!}{7!(46-7)!}=53524680$

We calculate the number of favorable combinations: for 1 red balls

$C_1^{12}\cdot C_7^{34}=12\cdot 1344904=16138848$

Therefore, the probability is

$P_1=\frac{16138848}{53524680}\\\\P_1=0.3$

We calculate the number of favorable combinations: for none red balls

$C_7^{34}=5379616$

Therefore, the probability is

$P_0=\frac{5379616}{53524680}\\\\P_0=0.1$

Therefore, the the probability that are at least 2 red balls is

$P=1-P_1-P_0\\\\P=1-0.3-0.1\\\\P=0.6$

c) We calculate the probability that are all withdrawn balls are the same color.

We calculate the number of possible combinations:

$C_7^{46}=\frac{46!}{7!(46-7)!}=53524680$

We calculate the number of favorable combinations:

$C_7^{12}+C_7^{16}+C_7^{18}=792+11440+31824=44056$

Therefore, the probability is

$P=\frac{44056}{53524680}\\\\P=0.0008$

d) We calculate the probability that are either exactly 3 red balls or exactly 3 blue balls are withdrawn.

Let X, event that exactly 3 red balls selected.

$P(X)=\frac{C_3^{12}\cdot C_4^{34}}{53524680}=0.57\\$

Let Y, event that exactly 3 blue balls selected.

$P(Y)=\frac{C_3^{16}\cdot C_4^{30}}{53524680}=0.29\\$

We have

$P(X\cap Y)=\frac{18\cdot C_3^{12} C_3^{16}}{53524680}=0.12$

Therefore, we get

$P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)\\\\P(X\cup Y)=0.57+0.29-0.12\\\\P(X\cup Y)=0.74$

8 0

3 years ago

I need the answer for 6

Gnoma [55]

ME AGAIN!!! You should just let me do your homework from now on lol.

So, you have 4 red, 3 green and 1 yellow. Add all of those numbers together, you should get 8. How many green did you start with? You started with 3. So your answer is either 3:8, 3/8, or 3 to 8.

P.S: I can see the page fine. It isn't blurry. Hope that helped you!

5 0

3 years ago

For two events A and B show that P (A∩B) ≥ P (A)+P (B)−1.

nordsb [41]

Answer:

<h3>For two events A and B show that P (A∩B) ≥ P (A)+P (B)−1.</h3>

By De morgan's law

$(A\cap B)^{c}=A^{c}\cup B^{c}\\\\P((A\cap B)^{c})=P(A^{c}\cup B^{c})\leq P(A^{c})+P(B^{c}) \\\\1-P(A\cap B)\leq P(A^{c})+P(B^{c}) \\\\1-P(A\cap B)\leq 1-P(A)+1-P(B)\\\\-P(A\cap B)\leq 1-P(A)-P(B)\\\\P(A\cap B)\geq P(A)+P(B)-1$