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ololo11 [35]
4 years ago
9

If θ is an acute angle and sin θ = 2 1 , then cos θ =

Mathematics
1 answer:
ss7ja [257]4 years ago
3 0
First of all i think theres a mistakein the question;
i think it is sin θ = \frac{1}{2}

and then sin 30° = \frac{1}{2}

thus θ= 30°\frac{ \sqrt{3} }{2}
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Pls help will give brainliest and 5star plssssss
ludmilkaskok [199]

2)

P(4,-4) -->(-4, 7)

4 - 8 = -4           -------->left 8

-4 + 11 = 7         -------->up 11

Answer: left 8;  up 11

3)

C(3,-1) , left 4 up 1

3 - 4 = -1    -------->left 4

-1 + 1 = 0   -------->up 1

a)

(x , y) -->(x - 4 , y +1)

C(3, -1) -->C'(-1 , 0)

b)

(x , y) --> (x - 4, y + 1); (-1 , 0)

7 0
3 years ago
Graph this line using the slope and y-intercept:<br> y = x + 9
prohojiy [21]

Answer:

Step-by-step explanation:

5 0
4 years ago
Nick wants to estimate the percentage of customers that would be satisfied with the company. He surveys 150 randomly selected cu
PilotLPTM [1.2K]

Answer:

The sample is the 150 randomly selected customers.

Step-by-step explanation:

When you survey a group of people, to gauge whaever you are interested, these people are what makes your sample.

In this question:

Nick surveys 150 randomly selected customers to determine whether or not they were satisfied.

So the sample is the 150 randomly selected customers.

4 0
3 years ago
An ectopic pregnancy is twice as likely to develop when the pregnant woman is a smoker as it iswhen she is a nonsmoker. If 32 pe
Dmitriy789 [7]

Answer:

48.48%

Step-by-step explanation:

Let's assume that there is a number N of women.

32% of these are smokers, then there are 0.32*N smokers

then 68% of these are non-smokers, then there are 0.68*N non-smokers.

Let's assume that the probability of having a ectopic pregnancy for a non-smoker is p (and the probability for a smoker will be 2*p)

Then the number of women with an ectopic pregnancy that are non-smokers is:

p*0.68*N

The number of women with an ectopic pregnancy that are smokers is:

2*p*0.32*N

Then the total number of women with an ectopic pregnancy will be:

p*0.68*N + 2*p*0.32*N

The percentage of women having an ectopic pregnancy that are smokers is equal to the quotient between the number of women with an ectopic pregnancy that are smokers and the total number of women with an ectopic pregnancy, all that times 100%.

The percentage is:

P = \frac{2*p*0.32*N}{p*0.68*N + 2*p*0.32*N} *100\%

Taking p and N as common factors, we get:

P = \frac{2*p*0.32*N}{p*0.68*N + 2*p*0.32*N} *100\% = \frac{N*p}{N*p} \frac{2*0.32}{0.68 + 2*0.32} *100\%

Then we get:

\frac{2*0.32}{0.68 + 2*0.32} *100\% = 48.48\%

8 0
3 years ago
Factor completely 4x^2 - 4x - 8
irinina [24]

Answer:

4(x+1) x (x-2)

Step-by-step explanation:

8 0
3 years ago
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