The bumper car ride at the state fair has 2 red cars, 2 green cars, and 4 blue cars. Devon is first in line for the

Three consecutive integers can be represented by the variable n, n+1 and n+2. If the sum of three consecutive integers is 57, wh

pashok25 [27]

You have some unknown integer $n$ , and you know that adding this and the next two integers, $n+1$ and $n+2$ , gives a total of 57.

This means

$n+(n+1)+(n+2)=57$

The task is to find all three unknown integers. Notice that if you know the value of $n$ , then you pretty much know the value of the other three integers.

To find $n$ , solve the equation above:

$n+(n+1)+(n+2)=3n+3=57\implies 3n=54\implies n=\dfrac{54}3=18$

So if 18 is the first integer, then others must be 19 and 20.

7 0

3 years ago

Read 2 more answers

What is the answer to 32 3/4 - 12 1/2 =

Natasha_Volkova [10]

Answer:

20.25 :)

Step-by-step explanation:

8 0

3 years ago

Given the equation -2x+3y=12, find the missing value in the ordered pair 3, <br><br> Noo linkss plz

astraxan [27]

Answer:

(3,6)

Step-by-step explanation:

-2x+3y=12

Let x = 3

-2(3) +3y = 12

-6+3y = 12

Add 6 to each side

-6+6 +3y = 12+6

3y = 18

Divide each side by 3

3y/3 = 18/3

y = 6

5 0

3 years ago

Great Escape bought a vintage bicycle for $3,000. They marked it up

slega [8]

Answer:

Solution,

CP of a bicycle=$3000

Marked up=40%

Now,

→40% of $3000

→40/100x$3000

→$1200

→$3000+$1200=$4200

Again,

Discount(d)=30%

→30%of $4200

→30/100x$4200

→$1260

Hence, the discount price of the bicycle is $1260

Step-by-step explanation:

4 0

2 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

4 years ago