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3 years ago

13

Use power series to solve (x + 1)y' + 3y = 0

1 answer:

enyata [817]3 years ago

6 0

X= -4 for solving x and y= 0 for solving y

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What is the solution set for the inequality |-3-5x|-8<-2

castortr0y [4]

$|-3-5x|-8$

6 0

3 years ago

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need it quick 13 points

motikmotik

Answer:

4 And -4 i think it depends

Step-by-step explanation:

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3 years ago

PLEASE HELP!! ABD and DBC are complementary angles. 70, what's the measurement of x?

Studentka2010 [4]

It is 45 because complementary means 90

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3 years ago

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The ages of a random sample of five university professors are 39, 54, 61, 72, and 59. Using this information, find a 99% confide

kondor19780726 [428]

Answer:

99% confidence interval for the population standard deviation = (74.97 , 635.20).

Step-by-step explanation:

We are given that the ages of a random sample of five university professors are 39, 54, 61, 72 and 59. Also, it is provided that the ages of university professors are normally distributed.

So, firstly the pivotal quantity for 99% confidence interval for the population standard deviation is given by;

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

where, s = sample standard deviation

$\sigma$ = population standard deviation

n = sample of university professors = 5

Also, $s^{2}$ = $\frac{\sum (X-\bar X)^{2} }{n-1}$ = 144.5

So, 99% confidence interval for population standard deviation, $\sigma$ is;

P(0.2070 < $\chi^{2} __5_-_1$ < 14.86) = 0.99 {As the table of $\chi^{2}$ at 4 degree of freedom

gives critical values of 0.2070 & 14.86}

P(0.2070 < $\frac{(n-1)s^{2} }{\sigma^{2} }$ < 14.86) = 0.99

P( $\frac{ 0.2070}{(n-1)s^{2} }$ < $\frac{1 }{\sigma^{2} }$ < $\frac{ 14.86}{(n-1)s^{2} }$ ) = 0.99

P( $\frac{ (n-1)s^{2}}{14.86 }$ < $\sigma^{2}$ < $\frac{ (n-1)s^{2}}{0.2070 }$ ) = 0.99

99% confidence interval for $\sigma^{2}$ = ( $\frac{ (n-1)s^{2}}{14.86 }$ , $\frac{ (n-1)s^{2}}{0.2070 }$ )

= ( $\frac{ (5-1) \times 144.5^{2}}{14.86 }$ , $\frac{ (5-1) \times 144.5^{2}}{0.2070 }$ )

= (5620.525 , 403483.092)

99% confidence interval for $\sigma$ = ( $\sqrt{5620.525}$ , $\sqrt{403483.092}$ )

= (74.97 , 635.20)

Therefore, 99% confidence interval for the population standard deviation of the ages of all professors at the university is (74.97 , 635.20).

8 0

4 years ago

How to write this numbers in word 76.68

Ad libitum [116K]

Answer:

seventy-six and sixty-eight hundreths

Step-by-step explanation:

5 0

3 years ago

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