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alexandr402 [8]
3 years ago
11

Find the value of sin(θ) for an angle θ in standard position with a terminal ray that passes through the point (-5, -12)

Mathematics
2 answers:
IceJOKER [234]3 years ago
8 0

Answer:

c. -12/13

Step-by-step explanation:

it will be the y-coordinate divided by the distance from the origin, or (-12) / \sqrt{5^2+12^2} = -12/13

netineya [11]3 years ago
6 0

Answer:

C. -12/13

Step-by-step explanation:

(-5,-12) is in the third quadrant

So sin would be negative

Hypotenuse = sqrt(5² + 12²)

sqrt(169)

13

sin(theta) = -12/13

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Amy is planning the seating arrangement for her wedding reception. Each round table can sit 12 guests. The head table can sit th
storchak [24]

Answer:

23 tables.

Step-by-step explanation:

282- 8 = 274     274 divided by 12 = 22.8  round to the nearest whole number which is 23.

7 0
3 years ago
Luisa recieved $108 for a necklace she sold on an online auction. The auction site took a fe of 10% of selling price. What was t
Vera_Pavlovna [14]
I believe it might be 107.9 after if that’s what your asking-like taking off 10% of $108.
3 0
3 years ago
Please help me out with this
Ivanshal [37]

Answer:

374.1

Step-by-step explanation:

Area of Hexagon = \frac{3\sqrt{3} }{2} * a^{2}

The ' a ' is the side length.

So now we just plug in the values

\frac{3\sqrt{3} }{2} *12^{2} = 374.12

3 0
3 years ago
From experience, it is known that on average 10% of welds performed by a particular welder are defective. if this welder is requ
bulgar [2K]
Binomial distribution can be used because the situation satisfies all the following conditions:1. Number of trials is known and remains constant (n)2. Each trial is Bernoulli (i.e. exactly two possible outcomes) (success/failure)3. Probability is known and remains constant throughout the trials (p)4. All trials are random and independent of the othersThe number of successes, x, is then given byP(x)=C(n,x)p^x(1-p)^{n-x}whereC(n,x)=\frac{n!}{x!(n-x)!}
Here we're given
p=0.10  [ success = defective ]
n=3

(a) x=0
P(x)=C(n,x)p^x(1-p)^{n-x}
=C(3,0)0.1^0(1-0.1)^{3-0}
=1(1)(0.729)
=0.729

(b) x=2
P(x)=C(n,x)p^x(1-p)^{n-x}
=C(3,2)0.1^2(1-0.1)^{3-2}
=3(0.01)(0.9)
=0.027

(c) x ≥ 2
P(x)=\sum_{x=2}^3C(n,x)p^x(1-p)^{n-x}
=P(2)+P(3)
=C(n,2)p^2(1-p)^{n-2}+C(n,3)p^3(1-p)^{n-3}
=C(3,2)0.1^2(1-0.1)^{3-2}+C(3,3)0.1^3(1-0.1)^{3-3}
=3(0.01)(0.9)+1(0.001)1
=0.027+0.001
=0.028


8 0
3 years ago
Johnathan ran 5 days this week. The most he ran in one day was 3.5 miles. Write an inequality that shows the distance johnathan
Mekhanik [1.2K]

<em><u>An inequality that shows the distance Johnathan could of ran any day this week is:</u></em>

x\leq 3.5

<em><u>Solution:</u></em>

Let "x" be the distance Johnathan can run any day of this week

Given that,

Johnathan ran 5 days this week. The most he ran in one day was 3.5 miles

Therefore,

Number of days ran = 5

The most he ran in 1 day = 3.5 miles

Thus, the maximum distance he ran in a week is given as:

distance = 5 \times 3.5 = 17.5

The maximum distance he ran in a week is 17.5 miles

If we let x be the distance he can run any day of this week then, we get a inequality as:

x\leq 3.5

If we let y be the total distance he can travel in a week then, we may express it as,

y\leq 17.5

8 0
3 years ago
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