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3 years ago

3.

Mathematics

1 answer:

seropon [69]3 years ago

4 0

Answer:

he needs 500 cm of ribbon

Step-by-step explanation:

pieces cut =10

size of pieces = 50 cm each .

50 × 10 = 500 answers

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How to calculate the percentage?

miv72 [106K]

$\: \: \: \: \:$

How to calculate percentage of a number. Use the percentage formula: P% * X = Y

Convert the problem to an equation using the percentage formula: P% * X = Y.
P is 10%, X is 150, so the equation is 10% * 150 = Y.
Convert 10% to a decimal by removing the percent sign and dividing by 100: 10/100 = 0.10.

Multiply by 100 and add a percentage sign
0.876 = 0.876 * 100 = 87.6%

<h2>hope it helps</h2>

7 0

3 years ago

Graph the given equation: y=2x+1

lions [1.4K]

So your slope is 2 and your Y-intercept is 1.

To graph it, start at 0 and go up to number 2.

8 0

3 years ago

Follow the steps and finish the solution.

Mazyrski [523]

Answer:

7 because 7x = 49 is 7 because 7 goes to itself 1 and to 49 7

4 0

3 years ago

18. You hear your little brother yelling from the other room about how his stupid phone has been

Savatey [412]

Answer:

40% I think

Step-by-step explanation:

if it was on 20% and it's only gone up 20% it would now be at 40% but it still only has gone up 20%

7 0

4 years ago

Let Y be a random variable with a density function given by

Neporo4naja [7]

From the given density function we find the distribution function,

$F_Y(y)=P(Y\le y)=\displaystyle\int_{-\infty}^y f_Y(t)\,\mathrm dt=\begin{cases}0&\text{for }y$

(a)

$F_{U_1}(u_1)=P(U_1\le u_1)=P(3Y\le u_1)=P\left(Y\le\dfrac{u_1}3\right)=F_Y\left(\dfrac{u_1}3\right)$

$\implies F_{U_1}(u_1)=\begin{cases}0&\text{for }u_1$

$\implies f_{U_1}(u_1)=\begin{cases}\frac{{u_1}^2}{18}&\text{for }-3\le u_1\le3\\0&\text{otherwise}\end{cases}$

(b)

$F_{U_2}(u_2)=P(3-Y\le u_2)=P(Y\ge3-u_2)=1-P(Y$

$\implies F_{U_2}(u_2)=\begin{cases}0&\text{for }u_2$

$\implies f_{U_2}(u_2)=\begin{cases}\frac32(u_2-3)^2&\text{for }2\le u_2\le4\\0&\text{otherwise}\end{cases}$

(c)

$F_{U_3}(u_3)=P(Y^2\le u_3)=P(-\sqrt{u_3}\le Y\le\sqrt{u_3})=F_Y(\sqrt{u_3})-F_y(\sqrt{u_3})$

$\implies F_{U_3}(u_3)=\begin{cases}0&\text{for }u_3$

$\implies f_{U_3}(u_3}=\begin{cases}\frac32\sqrt u&\text{for }0\le u\le1\\0&\text{otherwise}\end{cases}$

5 0

3 years ago