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34kurt
3 years ago
10

Given that cosθ = - and θ is in the second quadrant, find cscθ.

Mathematics
1 answer:
sdas [7]3 years ago
5 0
Since sin²x+cos²x=1, we can plug (-12/13) for cos(x) to get (-12/13)²+sin²x=1 
= 144/169+sin²x=1. Subtracting 144/169 from both sides, we get 25/169=sin²x. Square rooting both sides, we get 5/13 as sinx (since √25=5 and √169=13,as well as that it's in quadrant 2 - if it was in quadrant 3 or 4, it would be -5/13). Since cscx=1/sinx, we can plug (5/13) in for sinx to get 13/5 as our answer
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I need help with percent and tax
sasho [114]

Hello there

56•25/100=

1400÷100=

14$

56-14=42$

He will pay 42$

:D

6 0
3 years ago
A package of 10 batteries is checked to determine if there are any dead batteries. Four batteries are checked. If one or more ar
frutty [35]

Answer:

There is a probability of 76% of not selling the package if there are actually three dead batteries in the package.

Step-by-step explanation:

With a 10-units package of batteries with 3 dead batteries, the sampling can be modeled as a binomial random variable with:

  • n=4 (the amount of batteries picked for the sample).
  • p=3/10=0.3 (the proportion of dead batteries).
  • k≥1 (the amount of dead batteries in the sample needed to not sell the package).

The probability of having k dead batteries in the sample is:

P(x=k) = \dbinom{n}{k} p^{k}q^{n-k}

Then, the probability of having one or more dead batteries in the sample (k≥1) is:

P(x\geq1)=1-P(x=0)\\\\\\P(x=0) = \dbinom{4}{0} p^{0}q^{4}=1*1*0.7^4=0.2401\\\\\\P(x\geq1)=1-0.2401=0.7599\approx0.76

8 0
3 years ago
Find the circumfrance and the area of a circle with a 5 ft radius. Use the value 3.14
Crazy boy [7]

Answer:

Circumference is 2piR =31.4

Area is pir^2=78.5

Step-by-step explanation:

6 0
2 years ago
hello i really need help with this problem ive been stuck on it for 20 minutes so far i am willing to give 20 points if someone
Rasek [7]

Answer:

-9/20 with the given information i believe this is the answer

5 0
2 years ago
△QRS is rotated about point P to △Q′R′S′ .
Sedbober [7]

Answer:

As Given △QRS is rotated about point P to △Q′R′S′ .

As we know after rotation by any angle of pre-image neither the shape nor size of the image changes, the image after rotation is congruent to pre image.

Out of the options given

→The corresponding lengths, from the point of rotation, between the pre-image and the image are preserved.

→The corresponding angle measurements in each triangle between the pre-image and the image are preserved.

are true.



4 0
3 years ago
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