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3 years ago

If cos 67 degrees is close to 2/5 which is closest to the length of no

Mathematics

1 answer:

Juli2301 [7.4K]3 years ago

7 0

Answer:

40 cm

Step-by-step explanation:

In the right triangle PNO:

∠N = 90°

PO = 100cm (the hypotenuse of the triangle)

∠P = 23°

∠O = 90° - 23° = 67°

side NO is adjacent to ∠O

By definition:

cos(β) = adjacent/hypotenuse

Replacing with data and solving for NO:

cos(∠O) = NO/PO

2/5 = NO/100

NO = (2/5)*100

NO = 40

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Odd numbers between 18 and 82 that are multiples of 7

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3 years ago

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Use grid paper to find the median of the data. 9, 7, 2, 4, 3, 5,9,6,8,0,3, 8

Oksanka [162]

Answer:

5.5

Step-by-step explanation:

0,2,3,3,4,5,6,7,8,8,9,9

there are to middle numbers

add then together (5 and 6)

which makes 11

divide 11 by 2

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3 years ago

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

Find the scale factor of the smaller figure to the larger figure

4vir4ik [10]

Answer:

A) 3 : 5

Step-by-step explanation:

Scale factor = 24/40 = 3/5 = 3 : 5

5 0

3 years ago

I need this ASAP please!!!!

kipiarov [429]

Answer:

I think B is the answer ..........

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3 years ago

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