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mario62 [17]
3 years ago
7

Passing through (8,-6) and perpendicular to the line whose equation is y=1/2x+3

Mathematics
1 answer:
harina [27]3 years ago
3 0

Answer:

y=-2x+10

Step-by-step explanation:

y=1/2x+3 (8,-6)

y=mx+b

y=-2x+b

-6=-2(8)+b

-6=-16+b

+16 +16

10=b

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Given cosine of x is equal to negative square root of 3 over 2 comma what is the value of cos(x π)?
slamgirl [31]

Value of cos(x+π) = \bold{\frac{\sqrt{3} }{2}}

Given cosine of x = -\frac{\sqrt{3} }{2}

i.e., cos(x) =  -\frac{\sqrt{3} }{2}

Also cos(\frac{\pi}{6}) =  \frac{\sqrt{3} }{2}. So x = \frac{\pi}{6}.

We need to find cos(x+π) = cos(\frac{7\pi}{6}).

But its value is not easy to find. So we use trigonometric formulas.

We have the trigonometric formula, cos (x+π) = - cos(x) = -  -\frac{\sqrt{3} }{2} =  \bold{\frac{\sqrt{3} }{2}}

Cos(x+π) lies in the 3rd quadrant. So there cosine has negative values. That is why, cos (x+π) = - cos(x).

Learn more about Cos(x+π) trigonometric formula at brainly.com/question/12243383

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2 years ago
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SVETLANKA909090 [29]

Answer:

-4<1

Step-by-step explanation:u divide by 2 on both sides

7 0
3 years ago
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D.153.86in.............
Novay_Z [31]
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The breaking strengths of cables produced by a certain manufacturer have a mean, , of pounds, and a standard deviation of pounds
olchik [2.2K]

The question is incomplete. The complete question is :

The breaking strengths of cables produced by a certain manufacturer have a mean of 1900 pounds, and a standard deviation of 65 pounds. It is claimed that an improvement in the manufacturing process has increased the mean breaking strength. To evaluate this claim, 150 newly manufactured cables are randomly chosen and tested, and their mean breaking strength is found to be 1902 pounds. Assume that the population is normally distributed. Can we support, at the 0.01 level of significance, the claim that the mean breaking strength has increased?

Solution :

Given data :

Mean, μ = 1900

Standard deviation, σ = 65

Sample size, n = 150

Sample mean, $\overline x$ = 1902

Level of significance = 0.01

The hypothesis are :

$H_0 : \mu = 1900$

$H_1 : \mu > 1900$

Test statics :

We use the z test as the sample size is large and we know the population standard deviation.

$z=\frac{\overline x - \mu}{\sigma / \sqrt{n}}$

$z=\frac{1902-1900}{65 / \sqrt{150}}$

$z=\frac{2}{5.30723}$

$z=0.38$

Finding the p-value:

P-value = P(Z > z)

             = P(Z > 0.38)

             = 1 - P(Z < 0.38)

From the z table. we get

P(Z < 0.38) = 0.6480

Therefore,

P-value = 1 - P(Z < 0.38)

            = 1 - 0.6480

             = 0.3520

Decision :

If the p value is less than 0.01, then we reject the H_0, otherwise we fail to reject  H_0.

Since the value of p = 0.3520 > 0.01, the level of significance, then we fail to reject  H_0.

Conclusion :

At a significance level of 0.01, we have no sufficient evidence to support that the mean breaking strength has increased.

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horrorfan [7]
Sure where is the problem
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