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finlep [7]
2 years ago
10

Which are correct representation of the inequality minus 32X -5 less than 52 minus X select two options

Mathematics
1 answer:
Murljashka [212]2 years ago
5 0

Answer:

The value of X is more than equal to -1.83.

Step-by-step explanation:

We need to find the correct representation of the inequality 'minus 32X -5 less than 52 minus X'

LHS of the inequality will be : -32x-5

RHS of the inequality will be : (52-X)

So,

-32X-5\leq 52-X

Adding X both sides

-32X-5+X\leq 52-X+X\\\\-31X-5\leq 52

Adding 5 to both sides,

-31X-5+5\leq 52+5\\\\-31X\leq 57

Dividing both sides by -31.

X\geq \dfrac{-57}{31}\\\\X\geq  -1.83

So, the value of X is more than equal to -1.83.

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Well it depends...... (.5= 1/2)
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Type the correct answer in each box. Use numerals instead of words.
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Step-by-step explanation:

x+y=84

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y+6+y=84

2y+6=84

2y=84-6

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2 years ago
The sum of two numbers is 22. the difference is 6. what are the two numbers?
DaniilM [7]
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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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