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3 years ago

Tom and his brother caught 100 fish on the weeklog fishing trip. the total weight of the fish was 235 pounds.

1 answer:

5 0

The weight of 10 fish would be 23.5 pounds. If 100 fish is 235 pounds, then each fish is 2.35 pounds. 10 fish of that same weight would be 23.5 pounds.

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these are the means and standard deviations for samples of building heights in two different cities. city a city b mean: 120 ft

Kazeer [188]

The average building height and the standard deviation for two separate cities' samples are shown. The true statements are:

City A’s heights are more spread out than city B’s heights.

City A has a lower average height than city B.

<h3>What is average?</h3>

In daily life, an average is a number chosen to represent a group of numbers; it is often the sum of the numbers divided by the total number of numbers in the group (the arithmetic mean). The average of the integers 2, 3, 4, 7, and 9, for instance, is 5, which equals 25. An average could alternatively be another statistic like the median or mode, depending on the situation. The median is sometimes used in place of the mean since the mean would be unnaturally high if it included the personal incomes of a few billionaires. The median is the amount below which 50% of personal incomes fall and above which 50% of personal incomes rise.

To learn more about average ,visit:

brainly.com/question/15570401

#SPJ4

4 0

1 year ago

Write each equation in slope-intercept form of the equation of a line. Underline the slope and circle the y-intercept in each eq

Alik [6]

Answer:

see below:

Step-by-step explanation:

2y–6=0

a. slope intercept form using y = mx + b

y = 6/2

y = 3

b. slope: use the slope intercept form: y = mx + b

slope = m = 0

c. y-intercept = (0,3)

7 0

3 years ago

Suppose X, Y, and Z are random variables with the joint density function f(x, y, z) = Ce−(0.5x + 0.2y + 0.1z) if x ≥ 0, y ≥ 0, z

kompoz [17]

$f_{X,Y,Z}(x,y,z)=\begin{cases}Ce^{-(0.5x+0.2y+0.1z)}&\text{for }x\ge0,y\ge0,z\ge0\\0&\text{otherwise}\end{cases}$

is a proper joint density function if, over its support, $f$ is non-negative and the integral of $f$ is 1. The first condition is easily met as long as $C\ge0$ . To meet the second condition, we require

$\displaystyle\int_0^\infty\int_0^\infty\int_0^\infty f_{X,Y,Z}(x,y,z)\,\mathrm dx\,\mathrm dy\,\mathrm dz=100C=1\implies \boxed{C=0.01}$

b. Find the marginal joint density of $X$ and $Y$ by integrating the joint density with respect to $z$ :

$f_{X,Y}(x,y)=\displaystyle\int_0^\infty f_{X,Y,Z}(x,y,z)\,\mathrm dz=0.01e^{-(0.5x+0.2y)}\int_0^\infty e^{-0.1z}\,\mathrm dz$

$\implies f_{X,Y}(x,y)=\begin{cases}0.1e^{-(0.5x+0.2y)}&\text{for }x\ge0,y\ge0\\0&\text{otherwise}\end{cases}$

Then

$\displaystyle P(X\le1.375,Y\le1.5)=\int_0^{1.5}\int_0^{1.375}f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy$

$\approx\boxed{0.12886}$

c. This probability can be found by simply integrating the joint density:

$\displaystyle P(X\le1.375,Y\le1.5,Z\le1)=\int_0^1\int_0^{1.5}\int_0^{1.375}f_{X,Y,Z}(x,y,z)\,\mathrm dx\,\mathrm dy\,\mathrm dz$

$\approx\boxed{0.012262}$

7 0

3 years ago

Joseph borrowed a book from a library. The library charged a fixed rental for the book and a late fee for every day the book was

Paha777 [63]

The coefficient, 0.25, of the expression represents the late fee for every day the book was overdue.

And the constant, 2, represents the fixed rental fee.

The variable, x, represents the number of days Joseph had the book for.

Hope this helps :)

5 0

3 years ago

Lucio quiere repartir su colección de estampillas entre la mayor cantidad de amigos pero de manera que todos reciban la misma ca

Daniel [21]

Answer:

Lucio podrá armar 5 grupos iguales de estampillas.

Step-by-step explanation:

El problema plantea un caso de distribución mediante la aplicación del divisor común mayor a todos los números involucrados. Para ello, primero debemos analizar los divisores de cada uno de estos números:

-30: divisible por 2, 3, 5, 6, 10, 15 y 30.

-75: divisible por 3, 5, 15, 25 y 75.

-160: divisible por 2, 4, 5, 8, 10, 16, 20, 32, 40, 80 y 160.

Así, podemos ver que el divisor común mayor a los tres números es 5, ya que 30, 75 y 160 dan como resultado un número entero tras ser divididos por 5.

Entonces, para poder repartir equitativamente sus estampillas entre la mayor cantidad de amigos posibles, deberá repartir 6 estampillas de animales, 15 estampillas de flores y 32 estampillas de ciudades a 5 amigos. De esta manera, Lucio podrá armar 5 grupos iguales de estampillas.

4 0

3 years ago