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DedPeter [7]
3 years ago
13

Hi! I hope everyone has a great day! But can someone help me with this question? ☺️

Mathematics
1 answer:
yuradex [85]3 years ago
5 0

Answer:

We have the same problem

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Csc\theta -cot\theta =\frac{sin}{1+cos\theta }
Inga [223]

Answer:

Step-by-step explanation:

Left Hand Side

Change to sin(theta) and cos(theta)

csc(theta) = 1/sin(theta)

cot(theta) = cos(theta)/sin(theta)

1/sin(theta) - cos(theta)/sin(theta)                   Put over Sin(theta) Common denominator

[1 - cos(theta)] / sin(theta)                                Multiply numerator and denominator by 1 + cos(theta)

(1 - cos(theta)(1 + cos(theta) ) / sin(thata)*(1 + cos(theta))

(1 + cos(theta)(1 - cos(theta)) = 1 - cos^2(theta)

sin^2(theta) / (sin(theta)* ( 1 + cos(theta)

sin(theta) / (1 + cos(theta) )

Right hand Side.

See Above.

4 0
3 years ago
web based software company is interested in estimating the proportion of individuals who use the Firefox browser. In a sample of
nata0808 [166]
<h2>Answer with explanation:</h2>

Let \hat{p} denotes the sample proportion.

As per given , we have

n= 200

\hat{p}=\dfrac{30}{200}=0.15

Critical value for 99% confidence : z_{\alpha/2}=2.576

Confidence interval for population proportion :-

\hat{p}\pm z_{\alpha/2} \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\\\\=0.15\pm (2.576)(\sqrt{\dfrac{0.15(1-0.15)}{200}})\\\\\approx0.15\pm0.065\\\\=(0.15-0.065,\ 0.15+0.065)\\\\=(0.085,\ 0.215)

Hence, a 99% confidence interval for the proportion of all individuals that use Firefox: (0.085,\ 0.215)

The lower limit on the 99% confidence interval = 0.085

6 0
3 years ago
Look I’m dumb not a big brain person so I need help
Iteru [2.4K]

Answer:

the answer is 100 I promise

7 0
3 years ago
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The above is not a function
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2 years ago
Express the shaded part of the picture as a fraction A. 5 ∕9 B. 9 ∕5 C. 4 ∕9 D. 9 ∕4
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We need the picture to answer the question....
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