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fredd [130]
2 years ago
5

Which function has a domain of all real numbers

Mathematics
2 answers:
Hoochie [10]2 years ago
8 0

Answer: where are the numbers

Step-by-step explanation:

Pachacha [2.7K]2 years ago
4 0

Answer:

Most of the functions we have studied in Algebra I are defined for all real numbers. This domain is denoted . For example, the domain of f (x) = 2x + 5 is , because f (x) is defined for all real numbers x; that is, we can find f (x) for all real numbers x.

Step-by-step explanation:

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X=-6+y.x-3y=28. Iii fcugyvycuc
Leto [7]

Answer:

(-6+y)-3y=28

-2y-6=28

-2y=34

y=-17

Then,

x=-6-17

x=-23

Step-by-step explanation:

8 0
3 years ago
(-3/4m)-(1/2)=2+(1/4,)
Aleks [24]
(-3/4m) -(1/2)=2+(1/4m)
-(1/2)=2+m
m= -3/2
5 0
3 years ago
How many quarter pound buyers can george make out of 3 and 1/2 pounds of ground beef?
weeeeeb [17]
Divide the total amount by the portion size.

= 3 1/2 ÷ 1/4
convert 3 1/2 to improper fraction

= (2*3+1)/2 ÷ 1/4
= 7/2 ÷ 1/4

to divide a fraction, we multiply by the reciprocal/inverse of 1/4

= 7/2 * 4/1
multiply numerators, then multiply denominators

= (7*4)/(2*1)
= 28/2

= 14

ANSWER: 14

Hope this helps! :)
8 0
3 years ago
Find the vertex of: y=6(x-1)2-8<br> (x,y)
Pepsi [2]
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5 0
2 years ago
Suppose that from the past experience a professor knows that the test score of a student taking his final examination is a rando
DENIUS [597]

Answer:

n=13.167^2 =173.369 and if we round up to the nearest integer we got n =174

Step-by-step explanation:

Previous concepts

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Let X the random variable who represents the test score of a student taking his final examination. We know from the problem that the distribution for the random variable X is given by:

X\sim N(\mu =73,\sigma =10.5)

From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

Solution to the problem

We want to find the value of n that satisfy this condition:

P(71.5 < \bar X

And we can use the z score formula given by:

z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}

And we have this:

P(\frac{71.5-73}{\frac{10.5}{\sqrt{n}}} < Z

And we can express this like this:

P(-0.14286 \sqrt{n} < Z< 0.14286 \sqrt{n} )=0.94

And by properties of the normal distribution we can express this like this:

P(-0.14286 \sqrt{n} < Z< 0.14286 \sqrt{n} )=1-2P(Z

If we solve for P(Z we got:

P(Z

Now we can find a quantile on the normal standard distribution that accumulates 0.03 of the area on the left tail and this value is: z=-1.881

And using this we have this equality:

-1.881 = -0.14286 \sqrt{n}

If we solve for \sqrt{n} we got:

\sqrt{n} = \frac{-1.881}{-0.14286}=13.167

And then n=13.167^2 =173.369 and if we round up to the nearest integer we got n =174

6 0
3 years ago
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