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3 years ago

Please help Extra hundred points and brainleist

Mathematics

1 answer:

Sloan [31]3 years ago

4 0

Answer:

Step-by-step explanation:

It is A. The problem asks for the interval when the marlin's height is always increasing. Therefore, you want to look at the left half of the parabola, as the line is going into the peak. The right half is declining. The peak is at around 2 (on the x-axis), so you know that the ending interval number has to be 2. The marlin's height begins to go up starting at 0, so the interval is 0 <= x <= 2. Any number past two is a declining height, and the increase starts at 0.

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Find sin θ and cos θ if tan θ= 1/4 and sin 0>0.

Serga [27]

Answer:

$\displaystyle \sin(\theta)= \frac{\sqrt{17}}{17}$

$\displaystyle \cos(\theta)= \frac{4 \sqrt {17} }{ 17 }$

Step-by-step explanation:

We want to find sin(θ) and cos(θ) given that tan(θ) = 1/4 and sin(θ) > 0.

First, since tan(θ) and sin(θ) are both positive, cos(θ) must be positive as well.

Recall that tangent is the ratio of the opposite side to the adjacent side.

Therefore, the hypotenuse is:

$h=\sqrt{4^2+1^2}=\sqrt{16+1}=\sqrt{17}$

So, with respect to θ, the opposite side is 1, the adjacent is 4, and the hypotenuse is √17.

Then it follows that:

$\displaystyle \sin(\theta)=\frac{1}{\sqrt{17}} = \frac{\sqrt{17}}{17}$

And that:

$\displaystyle \cos(\theta)= \frac{4}{\sqrt{17}} = \frac{4 \sqrt {17} }{ 17 }$

5 0

3 years ago

How can you use random samples to solve real world problems

yaroslaw [1]

Random samples can provide more information on a population of people or things.

Hope I could be of help :)

8 0

3 years ago

Read 2 more answers

Of all customers purchasing automatic garage-door openers, 75% purchase Swedish model. Let X = the number among the next 15 purc

Lelechka [254]

Answer:

$P(X=k) = {15 \choose k} * 0.75^{k}*0.25^{15-k}$

For any integer k between 0 and 15, and 0 for other values of k.

$P(X>10) = 0.2252+ 0.2252+ 0.1559+0.0668+0.0134 = 0.6865$

c) P(6 ≤ X ≤ 10) = 0.2737

d) μ = 15*0.75 = 11.25. σ² = 11.25*0.25 = 2.8125

Step-by-step explanation:

X is a binomial random variable with parameters n = 15, p = 0.75. Therefore

$P(X=k) = {15 \choose k} * 0.75^{k}*0.25^{15-k}$

For any integer k between 0 and 15, and 0 for other values of k.

P(X>10) = P(X=11) + P(X=12)+ P(X=13)+P(X=14)+P(x=15)

$P(X=11) = {15 \choose 11} * 0.75^{11} * 0.25^4 = 0.2252$

$P(X=12) = {15 \choose 12} * 0.75^{12} * 0.25^3 = 0.2252$

$P(X=13) = {15 \choose 13} * 0.75^{13} * 0.25^2 = 0.1559$

$P(X=14) = {15 \choose 14} * 0.75^{14} * 0.25 = 0.0668$

$P(X=15) = {15 \choose 15} * 0.75^{15} = 0.0134$

Thus,

$P(X>10) = 0.2252+ 0.2252+ 0.1559+0.0668+0.0134 = 0.6865$

c) P(6 ≤ X ≤ 10) = P(X = 6) + P(X = 7) + P(X = 8) + P(X=9) + P(X=10)

$P(X=6) = {15 \choose 6} * 0.75^{6} * 0.25^9 = 0.0034$

$P(X=7) = {15 \choose 7} * 0.75^{7} * 0.25^8 = 0.0131$

$P(X=8) = {15 \choose 8} * 0.75^{8} * 0.25^7 = 0.0393$

$P(X=9) = {15 \choose 9} * 0.75^{9} * 0.25^6 = 0.0918$

$P(X=10) = {15 \choose 10} * 0.75^{10} * 0.25^{5} = 0.1652$

Thereofre,

$P(6 \leq X \leq 10) = 0.0034 + 0.0134 + 0.0393 + 0.0918 + 0.1652 = 0.2737$

d) μ = n*p = 15*0.75 = 11.25

σ² = np(1-p) = 11.25*0.25 = 2.8125

3 0

3 years ago

Evaluate sigma notation k=0 4 3k + 1

defon

The answer should be 126 or A.

4 0

3 years ago

-Name at least two different methods/activities to study material before a test.

nalin [4]

The methods/activities to study material before a test.

The SQ3R Method.
Retrieval Practice
Spaced Practice.

This is further explained below.

<h3>What is the SQ3R Method.?</h3>

The method is known as SQ3R. The SQ3R method is a reading comprehension strategy that teaches pupils how to recognize essential facts and keep material from their textbooks in their memories.

Generally, Learning may be improved and accelerated via the use of a technique known as retrieval practice, which involves actively recalling previously learned material. Recalling information on purpose compels us to "take out" our prior knowledge and investigate the basis of our current understanding.

In conclusion,

The SQ3R Method.
Retrieval Practice
Spaced Practice.