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Vika [28.1K]
3 years ago
7

Please help plz help

Mathematics
2 answers:
ch4aika [34]3 years ago
8 0
The area of circle is πr^2 which means 3.14(3)^2=28.26
kramer3 years ago
5 0

Answer: 530.93

Step-by-step explanation:

if this is not the answer I am sorry. plz tell me if I am right.

I am working on area of circles.

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A lawn with an area of 300 square feet needs 1/2 pound of grass seed. How many pounds of grass seed are needed for a lawn that i
lana66690 [7]
2 pounds.

60(20) = 1200 = area
If 300 square feet requires 1/2 pound, you can divide 1200 square feet by 300 square feet.

1200/300=4
4 times 1/2
4(1/2)=2

For every 300 square feet, you need half a pound. If you were to have 600 square feet, you would need a pound of grass seed. Therefore, for 1200 square feet, you need 2 pounds of grass seed.
4 0
3 years ago
Q # 28 please solve the equation
o-na [289]
The height of the pole = 129.3 ft

sin 36° = height of pole/ hypotenuse
hypotenuse = 220 ft

the height of the pole = 220 sin 36° = 129.3 ≈ 129 ft
4 0
3 years ago
Read 2 more answers
In this triangle, the product of sin B and tan C is (answer)
dmitriy555 [2]

Step-by-step explanation:

first one c/a

second one b/a

4 0
3 years ago
A commuter crosses one of three bridges, A, B, or C, to go home from work. The commuter crosses bridge A with probability 1/3, c
mihalych1998 [28]

Answer:

0.1333 = 13.33% probability that bridge B was used.

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Arrives home by 6 pm

Event B: Bridge B used.

Probability of arriving home by 6 pm:

75% of 1/3(Bridge A)

60% of 1/6(Bridge B)

80% of 1/2(Bridge C)

So

P(A) = 0.75*\frac{1}{3} + 0.6*\frac{1}{6} + 0.8*\frac{1}{2} = 0.75

Probability of arriving home by 6 pm using Bridge B:

60% of 1/6. So

P(A \cap B) = 0.6*\frac{1}{6} = 0.1

Find the probability that bridge B was used.

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.1}{0.75} = 0.1333

0.1333 = 13.33% probability that bridge B was used.

8 0
3 years ago
Writing to Explain Suppose you have a 100-mL cup, you can a 500-mL cup. List two different ways
Misha Larkins [42]
5 100ml cups and 1 500 ml cup


and 2 500 ml cups
4 0
3 years ago
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