All categories

3 years ago

What number represents 125.638 rounded to the nearest hundreth

Mathematics

2 answers:

Free_Kalibri [48]3 years ago

7 0

Answer:

125.64

Step-by-step explanation:

Nimfa-mama [501]3 years ago

4 0

Answer:

The answer is $125.64$

Step-by-step explanation:

hope this helps :)

You might be interested in

There are 9 students in a class: 7 boys and 2 girls.

Allisa [31]

Answer:

approximately 41.67%

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Tate has 5 more than twice as many pennies as Mia. If Mia has x pennies, which equation can be used to find how many pennies she

sladkih [1.3K]

Answer:

3x + 5 = 30

Step-by-step explanation:

Tate has 2x + 5 pennies and since Tate and Mia have 30 pennies all together you would add an x to the equation of how many pennies Tate has and set it equal to 30 making the equation 3x + 5 = 30

7 0

3 years ago

Read 2 more answers

A flat Samsung Tv 40" inches is 4 inches more than its width .Its area is 20 square inches .Find the width of the Flat Samsung T

SVETLANKA909090 [29]

FORMULA: AREA = LENGTH X WIDTH
with this we can proceed so the area is 20
the length is 40 inches and
answer is : width ----18.94---

5 0

3 years ago

1 + 1 - 1 = ?<br> For fun- <br> .........

coldgirl [10]

Answer:

1+1=2-1=1

Step-by-step explanation:

lol, this is the hardest question I've done today.

8 0

3 years ago

Read 2 more answers

Find the cdf F(x) associated with each of the following probability density functions. Sketch the graphs of f(x) and F(x).

wolverine [178]

Answer:

See steps below

Step-by-step explanation:

$\bf f(x)=3(1-x)^2\;(0$

$\bf F(x)=\int_{0}^{x}f(t)dt=\int_{0}^{x}3(1-t)^2dt=3\int_{0}^{x}(1-t)^2=1-(1-x)^3$

The cdf associated with f is

$\bf \boxed{F(x)=1-(1-x)^3}$ for 0<x<1

<h3>See picture 1 </h3>

The median is a point x such that

F(x) = ½

so, the median is

$\bf 1-(1-x)^3=1/2\rightarrow (1-x)^3=1/2\rightarrow \boxed{x=1-\sqrt[3]{2}}$

The 25th percentile equals the 1st quartile and is a point x such

F(x) = ¼

and the 25th percentile is

$\bf 1-(1-x)^3=1/4\rightarrow (1-x)^3=3/4\rightarrow \boxed{x=1-\sqrt[3]{3/4}}$

$\bf f(x)=\frac{1}{x^2}\;(1$

$\bf F(x)=\int_{1}^{x}f(t)dt=\int_{1}^{x}\frac{dt}{t^2}=1-\frac{1}{x}$

The cdf associated with f is

$\bf \boxed{F(x)=1-\frac{1}{x}}$ for x>1

<h3>See picture 2 </h3>

The median is

$\bf 1-\frac{1}{x}=\frac{1}{2}\rightarrow \boxed{x=2}$

The 25th percentile is

$\bf 1-\frac{1}{x}=\frac{1}{4}\rightarrow \boxed{x=4/3}$

f(x) = 1/3 for 0<x<1 or 2<x<4

$\bf F(x)=\int_{0}^{x}\frac{dt}{3}=\frac{x}{3}\;(0$

$\bf F(x)=\frac{1}{3}+\int_{2}^{x}\frac{dt}{3}=\frac{1}{3}+\frac{x-2}{3}=\frac{x-1}{3}\;(2$

The cdf associated with f is

$\bf F(x)=\frac{x}{3}$ for 0<x<1

$\bf F(x)=\frac{x-1}{3}$ for 2<x<4

<h3>See picture 3 </h3>

The median is

$\bf \frac{x-1}{3}=1/2\rightarrow x=1+3/2\rightarrow \boxed{x=5/2}$

The 25th percentile is

$\bf \frac{x}{3}=1/4\rightarrow \boxed{x=3/4}$

4 0

4 years ago