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Fynjy0 [20]
3 years ago
14

Giải pt lượng giác sau Cos x = -√3/2

Mathematics
1 answer:
Marat540 [252]3 years ago
6 0

Answer:

0

Step-by-step explanation:

Solution for cosx=-3/2 equation:

cos((3*pi)/5)*cos((3*pi)/20) = 0

(60*x^3)/(60*x^5) = 0

cos((3*pi)/5)*cos((3*pi)/20) = 0

cos((3*x)/5)*cos((3*x)/20) = 0

1.5/100 = 0

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If AB=6x-7 and BC=5x-3 find AC
luda_lava [24]

Answer:

AC= AB+BC

= 6x-7+ 5x-3

= 11x-10

4 0
3 years ago
The sum of one and the product of four and a number x
frez [133]
1+(4x) ??????????? I think so
7 0
3 years ago
Read 2 more answers
In a manufacturing process a random sample of 9 bolts manufactured has a mean length of 3 inches with a standard deviation of .3
Serggg [28]

Answer:

The correct option is (C) (2.769, 3.231).

Step-by-step explanation:

The confidence interval for mean when the standard deviation is not known is:

CI=\bar x\pm t_{\alpha/2, (n-1)}\frac{s}{\sqrt{n}}

Given:

\bar x = 3\\s=0.3\\n=9\\\alpha =1-0.95=0.05

Compute the critical value as follows:

t_{\alpha/2, (n-1)}=t_{0.05/2, (9-1)}=t_{0.025, 8}=2.31

**Use a <em>t</em>-table.

The 95% confidence interval for true mean length of the bolt is:

CI=\bar x\pm t_{\alpha/2, (n-1)}\frac{s}{\sqrt{n}}\\=3\pm 2.31\times \frac{0.30}{\sqrt{9}}\\ =3\pm 0.231\\=(2.769, 3.231)

Thus, the 95% confidence interval for true mean length of the bolt is (2.769, 3.231).

The correct option is (C).

7 0
3 years ago
How can we determine a slope?
zimovet [89]

slope formula: y2-y1/x2-x1

6 0
4 years ago
At a particular chess club, it is quite common for a chess game to take over two hours to complete. Suppose that the lengths of
Serga [27]

Answer:

A game would need to be at least 257.67 minutes long to qualify for the Endurance Board.

257.67 minutes = 257 minutes, 40 seconds = 4 hours, 17 minutes, 40 seconds.

Step-by-step explanation:

Games that are given special recognition on the Endurance Board are the games that last in the longest 1% of all games.

If X is the random variable that represents the time a chess game takes before it is completed.

X is said to be normally distributed with

Mean = μ = 153 minutes

Standard deviation = σ = 45 minutes

Let games that last the longest 1% of the time last for a minimum of x' minutes.

P(X > x') = 1% = 0.01

P(X ≤ x') = 1 - P(X > x') = 1 - 0.01

P(X ≤ x') = 0.99

Indicating that such games are longer than 99% of all chess games.

This is a normal distribution problem

Let the z-score for these type of longest games with a minimum duration of x' minutes be z'.

P(X ≤ x') = P(z ≤ z') = 0.99

From the normal distribution table, z' = 2.326

z-score of any value is given as the value minus the mean divided by the standard deviation.

z = (x - μ)/σ

So,

z' = (x' - μ)/σ

2.326 = (x' - 153)/45

x' = (2.326×45) + 153

x' = 104.67 + 153 = 257.67 minutes = 257 minutes, 40 seconds.

Hope this Helps!!!

7 0
3 years ago
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