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atroni [7]
3 years ago
12

Please help and i will give you the brainleist, please write the answer i notebook and send.​

Mathematics
1 answer:
Lina20 [59]3 years ago
6 0

Answer:

Step-by-step explanation:

Please see this link to get answer

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PLEASE HELPP and no pictures please i can't see them on a school computer. Ill appreciate a step by step explanation than you! I
babunello [35]

Answer:

wassup

Step-by-step explanation:

6 0
3 years ago
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Find the area of the shaded regions
iris [78.8K]

Answer:

buut

Step-by-step explanation:

6 0
3 years ago
Solve the equation.<br> <img src="https://tex.z-dn.net/?f=%5Csqrt%7By%5E%7B2%7D-20%2B100%20%7D%20%3Dy-10" id="TexFormula1" title
Lady bird [3.3K]

Answer:

No solution.

Step-by-step explanation:

So lets solve the square root.

Sqrt(y^2 - 20 + 100) = y - 10

Sqrt(y^2 + 80 = y - 10)

Solve time.

y^2 + 80 = y - 10 (We are squaring them)

y^2 + 80 = y^2 - 20y + 100

y^2 + 80 - y^2 = y^2 - 20y + 100 - y^2 (Subtract y^2 from both sides)

80 = -20y + 100

Flip it

-20y + 100 = 80

-20y + 100 - 100 = 80 - 100 (Subtract 100 from both sides)

-20y = -20

Now divide both sides by -20

-20y/-20 = -20/-20

y = 1

All you would do is plug in the values and you get...

9 \neq -9

This is false.

6 0
2 years ago
The Acme Company manufactures widgets. The distribution of widget weights is bell-shaped. The widget weights have a mean of 55 o
Usimov [2.4K]

Answer:

a) For this case and using the empirical rule we can find the limits in order to have 9% of the values:

\mu -2\sigma = 55 -2*6 =43

\mu +2\sigma = 55 +2*6 =67

95% of the widget weights lie between 43 and 67

b) For this case we know that 37 is 3 deviations above the mean and 67 2 deviations above the mean since within 3 deviation we have 99.7% of the data then the % below 37 would be (100-99.7)/2 = 0.15% and the percentage above 67 two deviations above the mean would be (100-95)/2 =2.5% and then we can find the percentage between 37 and 67 like this:

100 -0.15-2.5 = 97.85

c) We want to find the percentage above 49 and this value is 1 deviation below the mean so then this percentage would be (100-68)/2 = 16%

Step-by-step explanation:

For this case our random variable of interest for the weights is bell shaped and we know the following parameters.

\mu = 55, \sigma =6

We can see the illustration of the curve in the figure attached. We need to remember that from the empirical rule we have 68% of the values within one deviation from the mean, 95% of the data within 2 deviations and 99.7% of the values within 3 deviations from the mean.

Part a

For this case and using the empirical rule we can find the limits in order to have 9% of the values:

\mu -2\sigma = 55 -2*6 =43

\mu +2\sigma = 55 +2*6 =67

95% of the widget weights lie between 43 and 67

Part b

For this case we know that 37 is 3 deviations above the mean and 67 2 deviations above the mean since within 3 deviation we have 99.7% of the data then the % below 37 would be (100-99.7)/2 = 0.15% and the percentage above 67 two deviations above the mean would be (100-95)/2 =2.5% and then we can find the percentage between 37 and 67 like this:

100 -0.15-2.5 = 97.85

Part c

We want to find the percentage above 49 and this value is 1 deviation below the mean so then this percentage would be (100-68)/2 = 16%

4 0
4 years ago
Solve the given equation. (Enter your answers as a comma-separated list. Let k be any integer. Round terms to two decimal places
Ghella [55]

Answer:

That is the solutions are:

\frac{\pi}{3}+2\pi \cdot k,\frac{5\pi}{3}+2\pi \cdot k

Step-by-step explanation:

The period of \cos(x) or \sin(x) is 2\pi.

Let's look at the first rotation to see when \cos(\theta)=\frac{1}{2} happens.

This happens at \frac{\pi}{3} and also at \frac{5\pi}{3}. (Notice I just looked at the x-coordinates because that is what cosine is. Sine is the y-coordinate.)

Now to find the rest of the solutions we can just make full rotations either way to get back to those points .

That is the solutions are:

\frac{\pi}{3}+2\pi \cdot k

\frac{5\pi}{3}+2\pi \cdot k

7 0
4 years ago
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