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andreev551 [17]
3 years ago
6

I need to figure out which ones are irrational. please help

Mathematics
1 answer:
julsineya [31]3 years ago
3 0
Fatmagul neushe mode
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67

Step-by-step explanation:

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Which of the following expressions are factors of 25x^3−75x^2−9x+27?
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I am not sure but I think the correct answer is c

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This morning, Justin's car had 28.03 gallons of fuel. Now, 1.1 gallons are left. How much fuel did Justin use?
BARSIC [14]

Answer:

26.93 gallons were used

Step-by-step explanation:

Subtract the amount left from the amount started with

28.03

-  1.1

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26.93 gallons were used

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4 years ago
Midpoint of (-4,8) and (4,-3)
Alex73 [517]

Answer:

(0,5/2)

Step-by-step explanation:

(-4 , 8) ; (4 , -3)

Midpoint = (\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})\\\\=(\frac{-4 + 4}{2},\frac{8 - 3}{2})\\\\=(0 ,\frac{5}{2})

7 0
3 years ago
Let Y1,Y2, . . . ,Yn be a random sample from a normal distribution where the mean is 2 and the variance is 4. How large must n b
Marina86 [1]

Answer:

n= 60

Step-by-step explanation:

Hello!

You have Y₁, Y₂, ..., Yₙ random sample with a normal distribution: Y~N(μ;σ²)

μ= 2

σ²= 4

You need to calculate a sample size n so that (1.9 ≤ Y ≤2.1)= 0.99

To reach the sample size you need to work with the distribution of the sample mean (Y[bar]) because it is this distribution that is directly affected by the sample size.

Y[bar]~N(μ;σ²/n)

Under the sample mean distribution you have to use the standard normal:

Z=  Y[bar] - μ  ~N(0;1)

σ/√n

Now the asked interval is:

P(1.9 ≤ Y[bar] ≤2.1)= 0.99

The upper bond is 2.1

The lower bond is 1.9

The difference between the two bonds is the amplitude of the interval a=2.1-1.9= 0.2

And the probability included between these two bonds is 0.99

With this in mind you can rewite it as an interval for the sample mean:

Y[bar] + Z_{1-\alpha /2}*(σ/√n) - (Y[bar] + Z_{1-\alpha /2}*(σ/√n))= 0.2

Using the semiamplitude (d) of the interval you can easly calculate the required sample:

d= a/2= 0.2/2= 0.1

d= Z_{1-\alpha /2}*(σ/√n)

d* Z_{1-\alpha /2}= σ/√n

√n*(d* [tex]Z_{1-\alpha /2}[/tex)= σ

√n= σ/(d* [tex]Z_{1-\alpha /2}[/tex)

n= (σ/(d* [tex]Z_{1-\alpha /2}[/tex))²

n= (2/(0.1* 2.586))²

n= 59,81 ≅ 60

I hope it helps!

4 0
3 years ago
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