The speed limit near Marjorie's house is 30 miles per hour.

What is the approximate area of the shaded region? the shaded area is 28 A.168.56 square inches B.696.08 square inches C.740.

svp [43]

The answer is B its quite easy if you look at it

5 0

2 years ago

PLEASE HELP

Maurinko [17]

I believe this is the right way to write the function

N = [log A − log P] / log(1.0175)

N =[ log (1210.26) - log (1000)]/log (1.0175)

N =log [1210.26/1000] / log (1.0175) = log (1.21026)/log(1.0175) = 0.08288/0.00753 = 11

N = 11 months

4 0

2 years ago

If u(x) = x^5 – x^4 + x^2 and v(x) = –x^2, which expression is equivalent to (u/v)(x)?

Effectus [21]

Hello:
x^5 – x^4 + x^2 =x^2(x^3 - x^2 +1)
(u/v)(x)= x^2(x^3 - x^2 +1)/(-x^2)
(u/v)(x)= - (x^3 - x^2 +1) = -x^3+x^2-1 ( x : no zero)

7 0

3 years ago

The mean annual income for people in a certain city is 37 thousand dollars, with a standard deviation of 28 thousand dollars. A

Aloiza [94]

Answer:

$P( 31 < \bar X< 41)$

And we can ue the z score formula given by:

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

And using this formula we got for the limits:

$z = \frac{31-37}{\frac{28}{\sqrt{50}}}= -1.515$

$z = \frac{41-37}{\frac{28}{\sqrt{50}}}= 1.01$

So we want to find this probability:

$P(-1.515$

Step-by-step explanation:

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".

Solution to the problem

Let X the random variable that represent the annual income of a population, and for this case we know the following info:

$\mu=37$ and $\sigma=28$ and we are omitting the zeros from the thousand to simplify calculations

We select a sample size of n=50>30.

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".

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

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

And we want to find this probability:

$P( 31 < \bar X< 41)$

And we can ue the z score formula given by:

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

And using this formula we got for the limits:

$z = \frac{31-37}{\frac{28}{\sqrt{50}}}= -1.515$

$z = \frac{41-37}{\frac{28}{\sqrt{50}}}= 1.01$

So we want to find this probability:

$P(-1.515$

4 0

3 years ago

What is the length of the line?

Lelu [443]

6cm or (6,1)

If ur talking about vectors It’s (6,1)

6 0

1 year ago