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anastassius [24]
3 years ago
7

I NEED HELP!!!

Mathematics
1 answer:
oksano4ka [1.4K]3 years ago
6 0

Answer:

x= -2, y = -3

Step-by-step explanation:

1. substitute x = 4+2y in the first equation: \begin{bmatrix}2\left(4+2y\right)+y=-7\end{bmatrix}

simplify it : \begin{bmatrix}8+5y=-7\end{bmatrix}

2. isolate y in 8 + 5y = -7 --> 5y = -7 - 8 -->

3. solve for y:  5y = -15 --> y = -15/5 --> y= -3

4. solve for x: x = 4 + 2y

x = 4 + 2(-3)

x = 4 + (-6)

x = -2

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timofeeve [1]

Answer:

solution :

    x = 4

Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

                    5*(x-10)-(30-15*x)=0

Step by step solution :

Step  1  :

Equation at the end of step  1  :

 5 • (x - 10) -  (30 - 15x)  = 0

Step  2  :

Step  3  :

Pulling out like terms :

3.1     Pull out like factors :

  20x - 80  =   20 • (x - 4)

Equation at the end of step  3  :

 20 • (x - 4)  = 0

Step  4  :

Equations which are never true :

4.1      Solve :    20   =  0

Step-by-step explanation:

4 0
3 years ago
AJKL is similar to ARST. What is the value of x?
dlinn [17]

Answer:

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Step-by-step explanation:

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4 0
3 years ago
Read 2 more answers
Allen�s hummingbird (Selasphorus sasin ) has been studied by zoologist Bill Alther A small group of 15 Allen� s hummingbirds has
Bogdan [553]

Answer:

1) 80% CI: [3.04; 3.26]gr

d= 0.11

2) n= 28 hummingbirds

Step-by-step explanation:

Hello!

The study variable of this experiment is:

X: the weight of a hummingbird. (gr)

And it has a normal distribution, symbolically: X~N(μ;σ²)

And (I hope I got it correctly) its population standard deviation is σ= 0.33

There was a sample of n= 15 hummingbirds taken, its sample mean X[bar]= 3.15 gr

1) You need to construct an 80% Confidence Interval for the population mean of the hummingbird's weight.

Since the study variable has a normal distribution, you can use either the standard normal distribution or the Student's t distribution. Both are useful to estimate the population mean. Since the population standard variance is known, the best choice is the Standard normal.

Z= <u> X[bar] - μ </u>~ N(0;1)

       σ/√n

The formula for the interval is:

X[bar] ± Z_{1- \alpha /2} * (σ/√n)

Z_{1- \alpha /2}= Z_{0.90} = 1.28

3.15 ± 1.28 * (0.33/√15)

[3.04; 3.26]gr

The margin of error (d) of a confidence interval is hal its amplitude (a)

a= Upper bond - Lower bond

d= (Upper bond - Lower bond)/2

d= \frac{(3.26-3.04)}{2} = 0.11

2) You need to calculate a sample size for a 80% Confidence interval for the average weight of the hummingbirds with a margin of error of d= 0.08

As I said before, the margin of error is half the amplitude of the interval, the formula you use to estimate the population mean has the following structure:

"point estamator" ± "margin of error"

Then the margin of error is:

d= Z_{1- \alpha /2} * (σ/√n)

Now what you have to do is rewrite the formula based on the sample size

d= Z_{1- \alpha /2} * (σ/√n)

\frac{d}{Z_{1- \alpha /2}}= σ/√n

√n * \frac{d}{Z_{1- \alpha /2}}= σ

√n = σ * \frac{Z_1- \alpha /2}{d}

n = (σ * \frac{Z_1- \alpha /2}{d})²

n=  (0.33 * \frac{1.28}{0.08})²

n= 27.8784 ≅ 28 hummingbirds.

I hope it helps!

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