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2 years ago

Solve the system of equations

Mathematics

2 answers:

aleksandr82 [10.1K]2 years ago

8 0

Answer:

B) x=2, y=1

Step-by-step explanation:

x+3y=5

-x+6y=4

solve the equation

x=5-3y

-x+6y=4

substitute the value of x into an equation

-1(5-3y)+6y=4

multiply numbers in parenthesis by -1

-5-3y+6y=4

put -5 as positive and add 5 to 4 and put -3y as positive and add to 6y.

9y=9

divide both sides by 9

y=1

substitute the value of y into an equation

x=5-3•1

multiply 3 to 1

x= 5-3

subtract 5 to 3

x=2

------------

x=2

y=1

------------

ira [324]2 years ago

7 0

Answer:

B) x = 2, y = 1

Step-by-step explanation:

Add the two equations to eliminate the x-variable.

(x +3y) +(-x +6y) = (5) +(4)

9y = 9 . . . . . . .simplify

y = 1 . . . . . . . . divide by 9

You can substitute this value into either equation to find x.

x + 3(1) = 5

x = 2 . . . . . . . . subtract 3

The solution is (x, y) = (2, 1).

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A consumer products company is formulating a new shampoo and is interested in foam height (in mm). Foam height is approximately

Genrish500 [490]

Answer:

a) 0.057

b) 0.5234

c) 0.4766

Step-by-step explanation:

To find the p-value if the sample average is 185, we first compute the z-score associated to this value, we use the formula

$z=\frac{\bar x-\mu}{\sigma/\sqrt N}$

where

$\bar x=mean\; of\;the \;sample$

$\mu=mean\; established\; in\; H_0$

$\sigma=standard \; deviation$

N = size of the sample.

So,

$z=\frac{185-175}{20/\sqrt {10}}=1.5811$

$\boxed {z=1.5811}$

As the sample suggests that the real mean could be greater than the established in the null hypothesis, then we are interested in the area under the normal curve to the right of 1.5811 and this would be your p-value.

We compute the area of the normal curve for values to the right of 1.5811 either with a table or with a computer and find that this area is equal to 0.0569 = 0.057 rounded to 3 decimals.

So the p-value is

$\boxed {p=0.057}$

Since the z-score associated to an α value of 0.05 is 1.64 and the z-score of the alternative hypothesis is 1.5811 which is less than 1.64 (z critical), we cannot reject the null, so we are making a Type II error since 175 is not the true mean.

We can compute the probability of such an error following the next steps:

Step 1

Compute $\bar x_{critical}$

$1.64=z_{critical}=\frac{\bar x_{critical}-\mu_0}{\sigma/\sqrt{n}}$

$\frac{\bar x_{critical}-\mu_0}{\sigma/\sqrt{n}}=\frac{\bar x_{critical}-175}{6.3245}=1.64\Rightarrow \bar x_{critical}=185.3721$

So we would make a Type II error if our sample mean is less than 185.3721.

Step 2

Compute the probability that your sample mean is less than 185.3711

$P(\bar x < 185.3711)=P(z< \frac{185.3711-185}{6.3245})=P(z$

So, the probability of making a Type II error is 0.5234 = 52.34%

The power of a hypothesis test is 1 minus the probability of a Type II error. So, the power of the test is

1 - 0.5234 = 0.4766

3 0

2 years ago

an apple orchard contains a hundred and twelve trees the number of trees in each row is 2 less than twice the number of rows fin

san4es73 [151]

Let x be the number of rows and y be the number of trees per row.

x* y = 112
and y = 2x - 2
from first equation y = 112/x
so 112/x = 2x - 2
112 = 2x^2 - 2x
2x^2 - 2x - 112 = 0
x^2 - x - 56 = 0
(x - 8)(x + 7) = 0
x = 8 , - 7 (ignore the negative)
so the number of rows = 8
and number of trees per row = 112/8 = 14

Answer number of rows = 8 and number trees i each row = 14.

6 0

2 years ago

What is the area of a square if the base is 4.25ft and the height is 4.25 ft?

Oksi-84 [34.3K]

Answer:

18.0625 ft²

Step-by-step explanation:

4.25 × 4.25

18.0625 ft²

6 0

2 years ago

Read 2 more answers

A student placement center has requests from five students for interviews regarding employment with a particular consulting firm

balandron [24]

Answer:

a. P(both stat)=0,36

b. P(both math)=0,16

c. P( at least one sstatt)=0,84

d. P( at least on smath)=0,64

Step-by-step explanation:

Number of students= 5

Number of students math majors=2

Number of students statistics majors=3

a. Probability that both selected students are statistics majors

P(x)=3/5*3/5=0,36

b. Probability that both students are math majors

P(x)=2/5*2/5=0,16

c. Probability that at least one of the students selected is a statistics major

P(sstat)=3/5=0,6

P(not sstat)=2/5=0.4

P(sstat)=1-P(not sstat)∧n

P( at least on sstat)=1-0.4∧2=0.84

d. Probability that the selected students have different majors

P(smath)=2/5=0,4

P(not smath=2/5=0.6

P(smath)=1-P(not smath)∧n

P( at least on smath)=1-0.6∧2=0.64

4 0

2 years ago

The value of a $3000 computer decreases about 30% each year. write a function for the computers value V(t). Does the function re

Wittaler [7]

Answer:

Function represents decay.

$V(t)=3000(0.70)^t$

Step-by-step explanation:

Given that the value of a $3000 computer decreases about 30% each year. Now we need to write a function for the computers value V(t). then determine if the function represent growth or decay.

It clearly says that value decreases so that means function represents decay.

For decay we use formula:

$A=P(1-r)^t$

where P=initial value = $3000,

r= rate of decrease =30% = 0.30

t= number of years

A=V(t) = future value

so the required function is $V(t)=3000(1-0.30)^t$

or $V(t)=3000(0.70)^t$

7 0

2 years ago