Which equation has no solution?

You earn 5000 with a annual raise of 250, how much will you earn in 3 years?

Pachacha [2.7K]

Answer:

5750

Step-by-step explanation:

5000+250=5250

5250+350=5500

5500+250=5750

5 0

3 years ago

The average of a set of 5 numbers is 44. If the number 50 is added to the set, what is the new average?

tatuchka [14]

Answer:

45

Step-by-step explanation:

Um, If we say that all the numbers are 44, and now we add 50, then we will get 270. Find the mean of 270. We get 45.

5 0

4 years ago

2x + 2 y = -2<br> 3x - 2y = 12

Serggg [28]

Answer:

x=2 and y=-6 aka. (2,-6) which is the point of intersection

Step-by-step explanation:

Use elimination to cancel out y by adding the equations:

5x=10

x=2

Plug x=2 into one of the equations and find y:

2(5)+2y=-2

10+2y=-2

2y=-12

y=-6

Therefore, x=2 and y=-6

6 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

HELPPPPPPP ME PLEASE!!!!!

Snowcat [4.5K]

You can see that the two angles (38° and 7) form a line or 180°. So, angle 7 and 38° must add to 180°. So, angle 7 is:
$180-38 = 142$

Your answer is 142°.

4 0

4 years ago

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