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3 years ago

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Solution to this system of linear equation -6x=48-6y -4x=50-6y

1 answer:

MrRa [10]3 years ago

7 0

Answer:

$\large\boxed{x=1,\ y=6\to(1,\ 6)}$

Step-by-step explanation:

$\left\{\begin{array}{ccc}-6x=48-6y&\text{change the signs}\\-4x=50-6y\end{array}\right\\\underline{+\left\{\begin{array}{ccc}6x=-48+6y\\-4x=50-6y\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad2x=2\qquad\text{divide both sides by 2}\\.\qquad x=1\\\\\text{Put it ot the first equation:}\\-6(1)=48-6y\\-6=48-6y\qquad\text{subtract 48 from both sides}\\-54=-6y\qquad\text{divide both sides by (-6)}\\9=y\to y=6$

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Indicate the answer choice that best completes the statement or answers the question.

Tema [17]

Answer:

C

Step-by-step explanation:

m<2:

180 - 60 - 48 = 72

Vertical angle thm:

m<1 = 48

m<3:

180 - 62 - 48 = 70

6 0

3 years ago

complete the puzzle by unscrambling the letters below to reveal words from the vocabulary list at the beginning of the chapter

Marina CMI [18]

But what are we meant to unscramble?

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3 years ago

The mean preparation fee H&R Block charged retail customers in 2012 was $183 (The Wall Street Journal, March 7, 2012). Use t

astraxan [27]

Answer:

a)0.6192

b)0.7422

c)0.8904

d)at least 151 sample is needed for 95% probability that sample mean falls within 8$ of the population mean.

Step-by-step explanation:

Let z(p) be the z-statistic of the probability that the mean price for a sample is within the margin of error. Then

z(p)= $\frac{ME*\sqrt{N}}{s }$ where

Me is the margin of error from the mean

s is the standard deviation of the population

N is the sample size

a.

z(p)= $\frac{8*\sqrt{30}}{50 }$ ≈ 0.8764

by looking z-table corresponding p value is 1-0.3808=0.6192

b.

z(p)= $\frac{8*\sqrt{50}}{50 }$ ≈ 1.1314

by looking z-table corresponding p value is 1-0.2578=0.7422

c.

z(p)= $\frac{8*\sqrt{100}}{50 }$ ≈ 1.6

by looking z-table corresponding p value is 1-0.1096=0.8904

d.

Minimum required sample size for 0.95 probability is

N≥ $(\frac{z*s}{ME} )^2$ where

N is the sample size

z is the corresponding z-score in 95% probability (1.96)

s is the standard deviation (50)

ME is the margin of error (8)

then N≥ $(\frac{1.96*50}{8} )^2$ ≈150.6

Thus at least 151 sample is needed for 95% probability that sample mean falls within 8$ of the population mean.

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3 years ago

Your friend has $90 when he goes to the fair. He spends $5 to enter the fair and $15 on food. Rides at the fair cost $1.50 per r

Maksim231197 [3]

Answer: Option 'B' is correct.

Step-by-step explanation:

Since we have given that ,

Total money he takes while going to the fair = $90

Money he spends to enter the fair = $5

Money he spends on food =$15

Total he spent now is given by

$\$5+\$15=\$20$

Now, he spend on rides at the fair i.e. 1.50 per ride .

Let the number of rides be x

So, cost incurred on rides = 1.5x

So, the spending money can be expressed as

$20+1.50x$

Now, remaining money left to him after spending on x rides too is

$90-(1.50x+20)\\\\=90-1.50x-20\\\\=70-1.5x$

Let f(x) denotes the function used to determine the money he has left over after rides .

So it becomes

$f(x)=70-1.50x\\\\f(x)=-1.50x+70$

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3 years ago

Read 2 more answers

I need help please............

Anettt [7]

A) x=4
B) x= -3
C) x= -1/6
D) x=24

5 0

3 years ago