Una de las piezas de un monumento tiene forma crónica du altura es de 9cm y el area de su base de 10cm, su volumen es de?

Find the x-intercepts for the parabola defined

lana66690 [7]

Answer:

$x-intercept(s): (1,0),(-2,0)$

Step-by-step explanation:

5 0

3 years ago

Unit price $12 for 150lb bag

makvit [3.9K]

Answer:

For $1 its 12.5lbs

Step-by-step explanation:

6 0

3 years ago

List fractions and decimals in order from least to greatest 2.3, 24\5,2.6

vaieri [72.5K]

I can answer this for you its 4.8,2.6,and 2.3

4 0

3 years ago

The number of major earthquakes in a year is approximately normally distributed with a mean of 20.8 and a standard deviation of

SCORPION-xisa [38]

Answer:

a) 51.60% probability that in a given year there will be less than 21 earthquakes.

b) 49.35% probability that in a given year there will be between 18 and 23 earthquakes.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 20.8, \sigma = 4.5$

a) Find the probability that in a given year there will be less than 21 earthquakes.

This is the pvalue of Z when X = 21. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{21 - 20.8}{4.5}$

$Z = 0.04$

$Z = 0.04$ has a pvalue of 0.5160.

So there is a 51.60% probability that in a given year there will be less than 21 earthquakes.

b) Find the probability that in a given year there will be between 18 and 23 earthquakes.

This is the pvalue of Z when X = 23 subtracted by the pvalue of Z when X = 18. So:

X = 23

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{23 - 20.8}{4.5}$

$Z = 0.71$

$Z = 0.71$ has a pvalue of 0.7611

X = 18

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{18 - 20.8}{4.5}$

$Z = -0.62$

$Z = -0.62$ has a pvalue of 0.2676

So there is a 0.7611 - 0.2676 = 0.4935 = 49.35% probability that in a given year there will be between 18 and 23 earthquakes.

5 0

3 years ago

A store finds that its sales revenue changes at a rate given by S'(t) = −30t2 + 420t dollars per day where t is the number of da

allochka39001 [22]

Answer:

Step-by-step explanation:

Give the rate of change of sales revenue of a store modeled by the equation $S'(t)= -30t^{2} + 420t$ . The Total sales revenue function S(t) can be gotten by integrating the function given as shown;

$\int\limits {S'(t)} \, dt = \int\limits ({-30t^{2}+420t }) \, dt \\S(t) = \frac{-30t^{3} }{3}+\frac{420t^{2} }{2}\\ S(t)= -10t^{3} +210t^{2} \\$

a) The total sales for the first week after the campaign ends (t = 0 to t = 7) is expressed as shown;

$Given\ S(t) = -10t^{3} + 210t^{2}$

$S(0) = -10(0)^{3} + 210(0)^{2}\\S(0) = 0\\S(7) = -10(7)^{3} + 210(7)^{2}\\S(7) = -3430+10,290\\S(7) = 6,860$

Total sales = S(7) - S(0)

= 6,860 - 0

Total sales for the first week = $6,860

b) The total sales for the secondweek after the campaign ends (t = 7 to t = 14) is expressed as shown;

Total sales for the second week = S(14)-S(7)

Given S(7) = 6,860

To get S(14);

$S(14) = -10(14)^{3} + 210(14)^{2}\\S(14) = -27,440+41,160\\S(14) = 13,720$

The total sales for the second week after campaign ends = 13,720 - 6,860

= $6,860

8 0

4 years ago