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drek231 [11]
2 years ago
5

Jackson is attending a lacrosse game that is halfway through the first quarter.

Mathematics
1 answer:
shtirl [24]2 years ago
8 0

Answer:

4) 1/8

5) 56

6) 49

First shaded boxes = 4/7

Second shaded boxes = 1 out of 8 boxes are shaded = 1/8

Step-by-step explanation:

1/2x1/4=1/8

7X8=56

59-7=49

First shaded boxes = 4 out of 7 are shaded = 4/7

Second shaded boxes = 1 out of 8 boxes are shaded = 1/8

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Given that cos(theta) = 8/17 and that theta lies in Quadrant IV, what is the exact value of sin 2(theta)?
Deffense [45]

Answer:    -\frac{240}{289}

=========================================================

Explanation:

Use the pythagorean trig identity \sin^2(\theta)+\cos^2(\theta) = 1 and plug in the fact that \cos(\theta) = \frac{8}{17}\\\\

Isolating sine leads to \sin(\theta) = -\frac{15}{17}\\\\. I'm skipping the steps here, but let me know if you need to see them.

The result is negative because we're in quadrant 4, when y < 0 so it's when sine is negative.

Therefore,

\sin(2\theta) = 2\sin(\theta)\cos(\theta)\\\\\sin(2\theta) = 2*\left(-\frac{15}{17}\right)*\left(\frac{8}{17}\right)\\\\\sin(2\theta) = -\frac{240}{289}\\\\

6 0
1 year ago
Below is the graph of f ′(x), the derivative of f(x), and has x-intercepts at x = –3, x = 1 and x = 2. There are horizontal tang
mojhsa [17]

Answer:

C. f has a relative maximum at x = 1.

Step-by-step explanation:

A. False.  f(x) is concave down when f"(x) is negative.  f"(x) is the tangent slope of the graph, f'(x).  So f(x) is concave down between x = -1.5 and x = 1.5.

B. False.  f(x) is decreasing when f'(x) is negative.  So f(x) is decreasing in the intervals x < -3 and 1 < x < 2.

C. True.  f(x) has a relative maximum where f'(x) = 0 and changes from + to -.

8 0
2 years ago
Can someone help me whose good at geometry?
Pavel [41]
Actually I can help you in three of them as the number "3" is not confined between the arrays ND and NE

So:
It would be
angle END
or
angle DNE
or
angle N
6 0
3 years ago
The dwarves of the Grey Mountains wish to conduct a survey of their pick-axes in order to construct a 99% confidence interval ab
Dmitry_Shevchenko [17]

Answer:

The minimum sample size needed is 125.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

The margin of error is:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

For this problem, we have that:

\pi = 0.25

99% confidence level

So \alpha = 0.01, z is the value of Z that has a pvalue of 1 - \frac{0.01}{2} = 0.995, so Z = 2.575.

What minimum sample size would be necessary in order ensure a margin of error of 10 percentage points (or less) if they use the prior estimate that 25 percent of the pick-axes are in need of repair?

This minimum sample size is n.

n is found when M = 0.1

So

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.1 = 2.575\sqrt{\frac{0.25*0.75}{n}}

0.1\sqrt{n} = 2.575{0.25*0.75}

\sqrt{n} = \frac{2.575{0.25*0.75}}{0.1}

(\sqrt{n})^{2} = (\frac{2.575{0.25*0.75}}{0.1})^{2}

n = 124.32

Rounding up

The minimum sample size needed is 125.

5 0
3 years ago
How do you find the arctan 4/3 ? (Without a calculator)
Soloha48 [4]
If arctan 4/3 = the tangent of that angle will be 4/3 , opposite 4 over adjacent 3

Now, you only have to take the sin of that angle it's opposite 4 over hypotenuse 5. 

You will get that the answer is 4/5

hope this helps



7 0
2 years ago
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