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2 years ago

What is the solution to the system of equations?

Mathematics

1 answer:

vesna_86 [32]2 years ago

7 0

The correct answer is (10,-20), because the equation already gives you the correct x value. Then, from there you just plug 10 in for x in the y equation. Giving you a -20, :)

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Andres dog weighs 18 pounds. The vet says the breed of his dog typically weighs about 8 kg. How many kg is andres dog above the

Liono4ka [1.6K]

18-8=10

10*2.2=22

his dog is 22 lbs above the typical weight

7 0

3 years ago

A police car's siren operates at a frequency of 2,000 Hz. What frequency would a stationary listener observe if the police car i

SpyIntel [72]

we can use formula

$f_L=f_s(\frac{v_a-v_p}{v_a+v_p} )$

where

fL is listener frequency

fs is siren frequency

va is the speed of sound in air

vs is police car speed

so, we have

$f_s=2000Hz$

$v_a=345m/s$

$v_p=65mph$

now, we can change it into m/s

$v_p=65*\frac{1609}{3600}m/s$

$v_p=29.05139m/s$

now, we can plug it

$f_L=2000(\frac{345-29.05139}{345+29.05139} )$

$f_L=1689.33263Hz$ ..............Answer

3 0

3 years ago

!!20 POINTS!!

Sidana [21]

Answer:33.415 through 33.585 Or A

Step-by-step explanation:

You simply take 33.5 and add .085 to 33.5 get your answer.then subtract .085 from 33.5

6 0

2 years ago

Simplify the equation and show work!!!!

snow_tiger [21]

Step-by-step explanation:

This is the correct answer.

Even according to calculator and working.

So not quite sure why those options are incorrect?

4 0

2 years ago

In order to estimate the average time spent on the computer terminals per student at a local university, data were collected fro

VladimirAG [237]

Answer:

Margin of error for a 95% of confidence intervals is 0.261

Step-by-step explanation:

Step1:-

Sample n = 81 business students over a one-week period.

Given the population standard deviation is 1.2 hours

Confidence level of significance = 0.95

Zₐ = 1.96

Margin of error (M.E) = $\frac{Z_{\alpha }S.D }{\sqrt{n} }$

Given n=81 , σ =1.2 and Zₐ = 1.96

Step2:-

 $Margin of error (M.E) = \frac{Z_{\alpha }S.D }{\sqrt{n} }$ 

 $Margin of error (M.E) = \frac{1.96(1.2) }{\sqrt{81} }$ 

On calculating , we get

Margin of error = 0.261

Conclusion:-

Margin of error for a 95% of confidence intervals is 0.261

4 0

3 years ago