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4 years ago

4) Which step could be completed in order to prepare the system of equations below for the elimination method?

Mathematics

1 answer:

borishaifa [10]4 years ago

3 0

Answer:

Option C - Multiply the 1st equation by 2

Step-by-step explanation:

To find : Which step could be completed in order to prepare the system of equations below for the elimination method?

Solution :

1st equation: $5x-3y=10$

2nd equation: $8x+6y=15$

For elimination method,

We eliminate the y by multiplying 1st equation by 2 and add by 2nd equation.

i.e. 1st equation became $10x-6y=20$

Add with 2nd equation,

$10x-6y+8x+6y=20+15$

$18x=35$

$x=\frac{35}{18}$

Substitute in (1),

$5(\frac{35}{18})-3y=10$

$\frac{175}{18}-3y=10$

$3y=\frac{175}{18}-10$

$3y=\frac{175-180}{18}$

$3y=\frac{-5}{18}$

$y=\frac{-5}{54}$

Therefore, the step could be completed in order to prepare the system of equations below for the elimination method is option C.

You might be interested in

How many positive integers between 1000 and 9999 inclusive

bekas [8.4K]

Answer:

Step-by-step explanation:

first number is 1000-1+9=1008

9)1000(1

-------

-----

----

last number is 9999

9| 9999

---------

1111 |0

--------

9999=1008+(n-1)9

9999-1008=(n-1)9

n-1=8991/9=999

n=999+1=1000

first digit=1000

last digit=9999-1=9998

2 |9999

---------

|4999|1

9998=1000+(n-1)2

(n-1)2=9998-1000=8998

n-1=4499

n=4499=1=5000

c.not sure

total numbers=9000

9999=1000+(n-1)1

9999-1000=n-1

n=8999+1=9000

numbers divisible by 3=3000

first number=1002

last number=9999

9999=1002+(n-1)3

(n-1)3=9999-1002=8997

n-1=2999

n=2999+1=3000

numbers not divisible by 3=9000-3000=6000

numbers divisible by 5=1800

first number=1000

last number=9995

9995=1000+(n-1)5

(n-1)5=9995-1000=8995

n-1=1799

n=1799+1=1800

numbers divisible by 7=1286

7 | 1000

---------

| 142-6

1000-6+7=1001

7 | 9999

|---------

1428-3

9999-3=9996

first digit=1001

last digit=9996

9996=1001+(n-1)7

(n-1)7=9996-1001=8995

n-1=1285

n=1285+1=1286

numbers divisible by 35=257

first digit=1015

35 ) 1000 ( 28

----

300

280

------

---

1000-20+35=1015

35)9999(285

----

299

280

-----

199

175

----

last digit=9999-24=9975

9975=1015+(n-1)35

(n-1)35=9975-1015=8960

n-1=8960/35=256

n=257

reqd. numbers=1800+1286-257=3019

7 0

3 years ago

The percentage of body fat of a random sample of 36 men aged 20 to 29 found a sample mean of 14.42. Find a 95% confidence interv

Rina8888 [55]

Answer:

$14.42-1.96\frac{6.95}{\sqrt{36}}=12.150$

$14.42+ 1.96\frac{6.95}{\sqrt{36}}=16.690$

So on this case the 95% confidence interval would be given by (12.150;16.690)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=14.42$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma=6.95$ represent the population standard deviation

n=36 represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$