Find the circumference of the circle shown below.

Which equation represents a line that is parallel to the line y = - 4x +5?

Vikentia [17]

Y= -4x+3

slope is still the same so it would be parallel to that line. Only the y-intercept is different so it would just pass through a different point

6 0

3 years ago

Read 2 more answers

How to convert 2 and 4/5 into an improper fraction

zvonat [6]

Answer:

Step-by-step explanation:

2 = 4/2 . 4/5 can not be a improper fraction

4 0

2 years ago

Brian's kite is flying above a field at the end of a 68m string.If the angle of elevation to the kite measures 70 degrees,how hi

Dvinal [7]

Answer:

$h = 63.92\ m$

Step-by-step explanation:

Given:

Angle of elevation = 70°

Length of string = 68 m

We need to find the height of the kite.

Solution:

Requirement figure attached in file.

Where:

BC = 68 m

AC = h (Height of the kite)

∠ABC = 70°

Using Cosine rule to find the height of the kite.

$AC = BC\times sin(70)$

$h = 68\times 0.94$

$h = 63.92\ m$

Therefore, height of the kite from Brian's head $h = 63.92\ m$ .

4 0

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

(-2, 8) (9, 10) (1,7) (3, 9) (10, 12)FunctionNot a Function

lesya692 [45]

First of all, we need to know what a function is.

A function is a binary relation between two sets that

8 0

1 year ago