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4 years ago

What does it mean to solve for a variable when the solution is not a number?

1 answer:

3 0

Combine like terms on both sides of the equation. Isolate the x term by subtracting 2x from both sides. ... The solution x = 0 means that the value 0 satisfies the equation, so there is a solution. “No solution” means that there is no value, not even 0, which would satisfy the equation.

You might be interested in

1. Find the ratio of of 25cm and 1m .<br> 2. Find the product of (a + 4)(a − 4).

mr Goodwill [35]

Answer:

1..

solution....

here given

25cm

1m=100cm

now,

25/100=1/4

1:4

2....

(a+4)(a-4)

a^2-4^2

a^2-16

Step-by-step explanation:

hope it will help you

7 0

3 years ago

The heights of North American women are normally distributed with a mean of 64 inches and a standard deviation of 2inches.a. Wha

Charra [1.4K]

Answer:

a) $P(X>66)=P(Z>1)=1-P(Z$

b) $P(\bar X >66)=P(Z>2)=1-P(Z$

c) $P(\bar X >66)=P(Z>10)=1-P(Z$

Step-by-step explanation:

1) Previous concepts

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".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

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".

2) Part a

Let X the random variable that represent the heights of a population, and for this case we know the distribution for X is given by:

$X \sim N(64,2)$

Where $\mu=64$ and $\sigma=2$

We are interested on this probability

$P(X>66)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X>66)=P(\frac{X-\mu}{\sigma}>\frac{66-\mu}{\sigma})=P(Z>\frac{66-64}{2})=P(Z>1)$

And we can find this probability on this way:

$P(Z>1)=1-P(Z$

3) Part b

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

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

$P(\bar X >66)=P(Z>\frac{66-64}{\frac{2}{\sqrt{4}}}=2)$

And using a calculator, excel or the normal standard table we have that:

$P(Z>2)=1-P(Z$

4) Part c

$P(\bar X >66)=P(Z>\frac{66-64}{\frac{2}{\sqrt{100}}}=10)$

And using a calculator, excel or the normal standard table we have that:

$P(Z>10)=1-P(Z$

3 0

3 years ago

What is f(x) + f(x) + f(x)?

siniylev [52]

its 3f(x) i just answered it

5 0

4 years ago

The radius of a circle is decreasing at a rate of 6.56.56, point, 5 meters per minute. At a certain instant, the radius is 12121

Ilia_Sergeevich [38]

Answer:

The rate of change of the area of the circle is approximately $-490.09 \ m^2/min$ .

Step-by-step explanation:

Given:

$\frac{dr}{dt} = -6.5 \ m/min$

radius $r =12 \ m$

We need to find the rate of change of the area of the circle at that instant.

Solution:

Now we know that;

Area of the circle is given by π times square of the radius.

framing in equation form we get;

$A= \pi r^2$

to find the rate of change of the area of the circle at that instant we need to take the derivative on both side.

$\frac{dA}{dt}=\frac{d(\pi r^2)}{dt}\\\\\frac{dA}{dt}= 2\pi r \frac{dr}{dt}$

Substituting the given values we get;

$\frac{dA}{dt}= 2\times \pi \times 12 \times -6.5\\\\\frac{dA}{dt}\approx -490.09 \ m^2/min$

Hence The rate of change of the area of the circle is approximately $-490.09 \ m^2/min$ .

6 0

3 years ago

Circle A is shown. Tangents L M and N M intersect at point M outside of the circle. Arc L N is 270 degrees. In the diagram of ci

luda_lava [24]

Answer:

$\angle m=90^o$

Step-by-step explanation:

The picture of the question in the attached figure

we know that

The measurement of the external angle is the half-difference of the arches that comprise

$\angle m=\frac{1}{2}(major\ arc\ LN-minor\ arc\ LN)$

Remember that the sum of the major arc and a minor arc is equal to 360 degrees

we have

$major\ arc\ LN=270^o$ ----> is given

$minor\ arc\ LN=360^o-270^o=90^o$

substitute the values

$\angle m=\frac{1}{2}(270^o-90^o)=90^o$

5 0

4 years ago

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