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anygoal [31]
3 years ago
8

Solve the equation d − 5 9 = 3 9

Mathematics
1 answer:
Allushta [10]3 years ago
4 0

Answer:

D = 98

Step-by-step explanation:

if you add up 59 + 39 you get 98 meaning D is 98

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Find the ratio of the area of triangle XBY to the area of triangle ABC for the given measurements, if
WINSTONCH [101]

Answer: B. 1/4

Step-by-step explanation:BY/BC=2/4=1/2

 

Area ratio:

 

Area of triangle XBY / Area of triangle ABC = (BY/BC)²

=(1/2)²

=1/4

So the answer is B.  1/4

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3 years ago
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Write each recurring decimal as a fraction in simplest form.<br> 0.286
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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
Around which line would the following cross-section need to be revolved to create a sphere?
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Answer:

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Step-by-step explanation:

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