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damaskus [11]
2 years ago
7

Find the value of c 3-2(1 + 6c) = -35

Mathematics
1 answer:
Alex777 [14]2 years ago
7 0

Answer:

C = 3

Step-by-step explanation:

By using the distributive property, we multiply 1 and 6c inside the brackets by -2. This gives us 3 -2 - 12c = -35. Next, we subtract 2 from 3, giving us 1 - 12c = -35. We will now subtract 1 from both sides, leaving us with -12c = -36, or simply 12c = 36. Finally, we divide by 12, giving us C = 3. I hope this helps!

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sesenic [268]
The syntax for the IF statement is as follows:
=IF(condition, value if true, value if false)

therefore, we can enter the information from the problem:

=IF($B$9>=470000,35000,1000)

7 0
4 years ago
Find the rectangular coordinates of the point with the polar coordinates (-7, 5pi/3)
ELEN [110]
\left(7\cdot\cos \dfrac{5\pi}{3},7\cdot\sin\dfrac{5\pi}{3}\right)=\left(7\cdot\dfrac{1}{2},7\cdot\left(-\dfrac{\sqrt3}{2}\right)\right)=\\
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5 0
4 years ago
In triangle △ABC, ∠ABC=90°, BH is an altitude. Find the missing lengths.
shtirl [24]

Answer:

HC=\sqrt{3}

Step-by-step explanation:

Let HC=x, it is given that AH=3HC, then AH=3x.

Since, from the given figure, ΔABC is similar to ΔBHC and ΔABC is similar to ΔABH.

Therefore, ΔABH is similar to ΔBHC, hence using the similarity conditions,

\frac{HC}{BH}=\frac{BH}{AH}

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7 0
4 years ago
The amount of lateral expansion (mils) was determined for a sample of n = 7 pulsed-power gas metal arc welds used in LNG ship co
zvonat [6]

Answer:

95% Confidence interval for σ2 and for σ is (3.33 , 38.85) and (1.82 , 6.23) respectively.

Step-by-step explanation:

We are given that the amount of lateral expansion (mils) was determined for a sample of n = 7 pulsed-power gas metal arc welds used in LNG ship containment tanks. The resulting sample standard deviation was s = 2.83 mils.

Assuming data follows normal distribution.

So, firstly the pivotal quantity for 95% confidence interval for the population variance is given by;

        P.Q. = \frac{(n-1)s^{2} }{\sigma^{2} } ~ \chi^{2} __n_-_1

where, s = sample standard deviation = 2.83 mils

          \sigma^{2} = population variance

           \sigma = population standard deviation

           n = sample size = 7

<em>So, 95% confidence interval for population variance, </em>\sigma^{2} <em>is;</em>

P(1.237 < \chi^{2} __n_-_1 < 14.45) = 0.95 {As the table of at 6 degree of freedom

                                                     gives critical values of 1.237 & 14.45}

P(1.237 < \frac{(n-1)s^{2} }{\sigma^{2} } < 14.45) = 0.95

P( \frac{ 1.237}{(n-1)s^{2} } < \frac{1 }{\sigma^{2} } < \frac{ 14.45}{(n-1)s^{2} } ) = 0.95

P( \frac{ (n-1)s^{2}}{14.45 } < \sigma^{2} < \frac{ (n-1)s^{2}}{1.237 } ) = 0.95

95% confidence interval for \sigma^{2} = ( \frac{ (n-1)s^{2}}{14.45 }  , \frac{ (n-1)s^{2}}{1.237 }  )

                                                  = ( \frac{ (7-1) \times 2.83^{2}}{14.45 } , \frac{ (7-1) \times 2.83^{2}}{1.237 } )

                                                  = (3.33 , 38.85)

95% C.I. for population standard deviation, \sigma  = ( \sqrt{3.33} , \sqrt{38.85} )

                                                                            = (1.82 , 6.23)

Therefore, 95% confidence interval for the population variance (σ2) and population standard deviation (σ) are (3.33 , 38.85) and (1.82 , 6.23) respectively.

7 0
3 years ago
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Soloha48 [4]

Answer:

d=-6

Step-by-step explanation:

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