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4 years ago

Have many solutions does -4x + 6 + 3x = 12

Mathematics

2 answers:

GrogVix [38]4 years ago

4 0

Answer:

A) One solution

B) No solution

C) Infinite number of solutions

D) One solution

E) Infinite number of solutions

Step-by-step explanation:

There are 5 equations to solve:

A) - 4 x + 6 + 3 x = -12

combining like terms and solving for x gives:

- x + 6 = - 12

- x = -18

x = 18

which is a unique (ONE) solution

B) 5 (2 x - 2) = 10 x + 10

distributive property gives us: 10 x - 10 = 10 x + 10

then subtract 10 x from both sides and we get: -10 = 10 which is an absurd, therefore this equation has NO SOLUTION.

C) 6 x + 2 x - 8 = 8 (x - 1)

Combining like terms and using distributive property, we get:

8 x - 8 = 8 x - 8

which is an identity, So this is true for all possible x values, then the are INFINITE NUMBER OF SOLUTIONS

D) 2 x + 50 = 90

solving for x gives:

2 x = 40

x = 40/2

x = 20

which is a unique (ONE) solution.

E) 2 (x - 4) = 2 x + 10 - 18

Using distributive property and combining like terms we get:

2 x - 8 = 2 x -8

This is an identity which will give true for all real x-values. Therefore it has INFINITE NUMBER OF SOLUTIONS.

LUCKY_DIMON [66]4 years ago

3 0

Answer:As there are 2 equations and two unknowns there is 1 solution

Step-by-step explanation:

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A Galapagos tortoise can 1/5 mile in an hour. How many hours would it take the tortoise to walk 1/6?

polet [3.4K]

Answer:

5/6 hours

Step-by-step explanation:

Create a proportion where x is the number of hours it will take to walk 1/6 of a mile:

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Cross multiply and solve for x:

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3 years ago

Which equation is a point slope form equation for line AB ?

Jobisdone [24]

For this case we have that by definition, the equation of a line of the point-slope form is given by:

$y-y_ {0} = m (x-x_ {0})$

Where:

m: It's the slope

$(x_ {0}, y_ {0}):$ It is a point through which the line passes

To find the slope, we need two points through which the line passes, observing the image we have:

$(x_ {1}, y_ {1}): (1,6)\\(x_ {2}, y_ {2}): (5, -2)\\m = \frac {y_ {2} -y_ {1}} {x_ {2} -x_ {1}} = \frac {-2-6} {5-1} = \frac {-8} {4} = -2$

Thus, the equation is of the form:

$y-y_ {0} = - 2 (x-x_ {0})$

We choose a point:

$(x_{0}, y_ {0}) :( 5, -2)$

Finally, the equation is:

$y - (- 2) = - 2 (x-5)\\y + 2 = -2 (x-5)$

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$y + 2 = -2 (x-5)$

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3 years ago

The Insurance Institute reports that the mean amount of life insurance per household in the US is $110,000. This follows a norma

nata0808 [166]

Answer:

a) $\sigma_{\bar X} = \frac{\sigma}{\sqrt{n}}= \frac{40000}{\sqrt{50}}= 5656.85$

b) Since the distribution for X is normal then we know that the distribution for the sample mean $\bar X$ is given by:

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

c) $P( \bar X >112000) = P(Z>\frac{112000-110000}{\frac{40000}{\sqrt{50}}}) = P(Z>0.354)$

And we can use the complement rule and we got:

$P(Z>0.354) = 1-P(Z$

d) $P( \bar X >100000) = P(Z>\frac{100000-110000}{\frac{40000}{\sqrt{50}}}) = P(Z>-1.768)$

And we can use the complement rule and we got:

$P(Z>-1.768) = 1-P(Z$

e) $P(100000< \bar X$

And we can use the complement rule and we got:

$P(-1.768$

Step-by-step explanation:

a. If we select a random sample of 50 households, what is the standard error of the mean?

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".

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

$X \sim N(110000,40000)$

Where $\mu=110000$ and $\sigma=40000$

If we select a sample size of n =35 the standard error is given by:

$\sigma_{\bar X} = \frac{\sigma}{\sqrt{n}}= \frac{40000}{\sqrt{50}}= 5656.85$

b. What is the expected shape of the distribution of the sample mean?

Since the distribution for X is normal then we know that the distribution for the sample mean $\bar X$ is given by:

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

c. What is the likelihood of selecting a sample with a mean of at least $112,000?

For this case we want this probability:

$P(X > 112000)$

And we can use the z score given by:

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

And replacing we got:

$P( \bar X >112000) = P(Z>\frac{112000-110000}{\frac{40000}{\sqrt{50}}}) = P(Z>0.354)$

And we can use the complement rule and we got:

$P(Z>0.354) = 1-P(Z$

d. What is the likelihood of selecting a sample with a mean of more than $100,000?

For this case we want this probability:

$P(X > 100000)$

And we can use the z score given by:

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

And replacing we got:

$P( \bar X >100000) = P(Z>\frac{100000-110000}{\frac{40000}{\sqrt{50}}}) = P(Z>-1.768)$

And we can use the complement rule and we got:

$P(Z>-1.768) = 1-P(Z$

e. Find the likelihood of selecting a sample with a mean of more than $100,000 but less than $112,000

For this case we want this probability:

$P(100000$

And we can use the z score given by:

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

And replacing we got:

$P(100000< \bar X$

And we can use the complement rule and we got:

$P(-1.768$

8 0

3 years ago