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3 years ago

What is the solution of the system?

Mathematics

2 answers:

nikitadnepr [17]3 years ago

8 0

Answer:

The World is now confused

Step-by-step explanation:

Ivenika [448]3 years ago

3 0

Answer:

(1, 2 )

Step-by-step explanation:

The solution to a system of equations given graphically is at the point of intersection of the 2 lines.

The lines intersect at (1, 2 ) ← solution

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matrenka [14]

Answer:

I think its 14 students

Step-by-step explanation:

40%x35=14

7 0

3 years ago

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Suppose you buy a 1.25 pound package f ham at $5.20 per pound what fraction of a pound did you buy

Andrew [12]

1 1/4

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8 0

3 years ago

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Anvisha [2.4K]

Heberto could go 14 blocks East and 14 blocks south, so the last answer

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3 years ago

Solve x2 − 12x + 5 = 0 using the completing-the-square method. x = six plus or minus the square root of five x = negative six pl

Andru [333]

we are given

$x^2-12x+5=0$

we have to solve it by completing square method

step-1: Move 5 on right side

$x^2-12x+5-5=0-5$

$x^2-12x=-5$

step-2: Break middle term

$x^2-2*6*x=-5$

step-3: Add 6^2 both sides

$x^2-2*6*x+(6)^2=-5+(6)^2$

$(x-6)^2=31$

step-3: Solve for x

$\sqrt{(x-6)^2} =\sqrt{31}$

we wil get two values

First value is

$x-6=\sqrt{31}$

add both sides 6

$x-6+6=6+\sqrt{31}$

$x=6+\sqrt{31}$

Second value is

$x-6=-\sqrt{31}$

add both sides 6

$x-6+6=6-\sqrt{31}$

$x=6-\sqrt{31}$

so, solutions are

$x=6+\sqrt{31}$

$x=6-\sqrt{31}$ .............Answer

5 0

3 years ago

When obtaining a confidence interval for a population mean in the case of a finite population of size N and a sample size n whic

Elanso [62]

Answer:

95% Confidence interval for the mean

$142.8 \leq\mu\leq158.4$

Step-by-step explanation:

We have to calculate a 95% confidence interval for the mean of a finite population.

The error is multiplied by the following finite population correction factor:

$cf=\sqrt{\frac{N-n}{N-1} }$

The standard deviation can be estimated as

$\sigma=\frac{s}{\sqrt{n}} \sqrt{\frac{N-n}{N-1} } =\frac{24.4}{\sqrt{32} }* \sqrt{\frac{200-32}{200-1} }=3.963$

The 95% confidence interval has a z value of 1.96, so it becomes:

$M-z*\sigma_c\leq\mu\leq M+z*\sigma_c\\\\150.6-1.96*3.963\leq\mu\leq 150.6+1.96*3.963\\\\ 142.8 \leq\mu\leq 158.4$

5 0

3 years ago