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3 years ago

Can someone help me with this problem? #32

Mathematics

1 answer:

Masja [62]3 years ago

8 0

Answer:

answer is A but im sure you've gotten it by now

Step-by-step explanation:

You might be interested in

The Jackson's dinner cost $125. They left $21.25 as a tip. What was the percentage of the tip? Please explain and help me! THANK

fiasKO [112]

Answer:

17%

Step-by-step explanation:

to find the percentage of something, you have to fill out the following formula:

x/100 = 21.25/125. Basically that is saying that some percent out of 100 is equal to 21.25 out of 125. To solve that, you have to cross multiply:

125x = 2125

then you have to divide, to single out x:

the answer is 17

so it is 17%

3 0

3 years ago

Read 2 more answers

Find the function y1 of t which is the solution of

Harrizon [31]

Answer:

a) y₁(t) = (2.5e⁰•⁸ᵗ - 2.5e⁰•⁴ᵗ)

b) y₂(t) = (2e⁰•⁴ᵗ - e⁰•⁸ᵗ)

c) W(y₁,y₂) = -e¹•²ᵗ

Step-by-step explanation:

To solve, 25y" - 30y' + 8y = 0

With conditions, y₁'(0) = 1 and y₁(0) = 0 for y₁

And y₂'(0) = 0 and y₂(0) = 1 for y₂

Solving for y₁ first, using Laplace transforms.

Note:

L(y") = s² Y(s) - sy(0) - y'(0)

L(y') = s Y(s) - y(0)

L(y₁") = s² Y(s) - sy₁(0) - y'₁(0) = s² Y(s) - 0 - 1 = s² Y(s) - 1

L(y₁') = s Y(s) - y₁(0) = s Y(s)

L [25y" - 30y' + 8y] = L (0)

25(s²Y(s) - 1) - 30(sY(s)) + 8Y(s) = 0

25s²Y(s) - 25 - 30sY(s) + 8Y(s) = 0

Y(s) [25s² - 30s + 8] - 25 = 0

Y(s) [25s² - 30s + 8] = 25

Y(s) = (25)/[25s² - 30s + 8]

Y(s) = (25)/[(5s-4)(5s-2)]

To continue, we need to resolve 1/[(5s-4)(5s-2)] into partial fractions.

The calculation is continued on the attached images.

8 0

4 years ago

Write a scenario that could be represented by this equation. 3/4x=12

taurus [48]

If a car wants to go a total distance of 12 miles and can go 3/4 of a mile every hour, then how many hours will it take the car to travel the entire distance?

5 0

3 years ago

Solve for t: if a≠−b t + <img src="https://tex.z-dn.net/?f=%5Cfrac%7Bat%7D%7Bb%7D" id="TexFormula1" title="\frac{at}{b}" alt

prohojiy [21]

Answer:

$t=\frac{b}{b+a}$

Step-by-step explanation:

we have

$t+\frac{at}{b}=1$

Solve for t

That means ----> Isolate the variable t

Multiply by b both sides to remove the fraction

$bt+at=b$

Factor the variable t in the left side

$t[b+a]=b$

Divide by (b+a) both sides

$t=\frac{b}{b+a}$

4 0

4 years ago

Assume a jar has five red marbles and three black marbles. Draw out two marbles with and without replacement. Find the requested

Doss [256]

Answer:

For probabilities with replacement

$P(2\ Red) = \frac{25}{64}$

$P(2\ Black) = \frac{9}{64}$

$P(1\ Red\ and\ 1\ Black) = \frac{15}{32}$

$P(1st\ Red\ and\ 2nd\ Black) = \frac{15}{64}$

For probabilities without replacement

$P(2\ Red) = \frac{5}{14}$

$P(2\ Black) = \frac{3}{28}$

$P(1\ Red\ and\ 1\ Black) = \frac{15}{28}$

$P(1st\ Red\ and\ 2nd\ Black) = \frac{15}{56}$

Step-by-step explanation:

Given

$Marbles = 8$

$Red = 5$

$Black = 3$

For probabilities with replacement

(a) P(2 Red)

This is calculated as:

$P(2\ Red) = P(Red)\ and\ P(Red)$

$P(2\ Red) = P(Red)\ *\ P(Red)$

So, we have:

$P(2\ Red) = \frac{n(Red)}{Total} \ *\ \frac{n(Red)}{Total}\\$

$P(2\ Red) = \frac{5}{8} * \frac{5}{8}$

$P(2\ Red) = \frac{25}{64}$

(b) P(2 Black)

This is calculated as:

$P(2\ Black) = P(Black)\ and\ P(Black)$

$P(2\ Black) = P(Black)\ *\ P(Black)$

So, we have:

$P(2\ Black) = \frac{n(Black)}{Total}\ *\ \frac{n(Black)}{Total}$

$P(2\ Black) = \frac{3}{8}\ *\ \frac{3}{8}$

$P(2\ Black) = \frac{9}{64}$

(c) P(1 Red and 1 Black)

This is calculated as:

$P(1\ Red\ and\ 1\ Black) = [P(Red)\ and\ P(Black)]\ or\ [P(Black)\ and\ P(Red)]$

$P(1\ Red\ and\ 1\ Black) = [P(Red)\ *\ P(Black)]\ +\ [P(Black)\ *\ P(Red)]$

$P(1\ Red\ and\ 1\ Black) = 2[P(Red)\ *\ P(Black)]$

So, we have:

$P(1\ Red\ and\ 1\ Black) = 2*[\frac{5}{8} *\frac{3}{8}]$

$P(1\ Red\ and\ 1\ Black) = 2*\frac{15}{64}$

$P(1\ Red\ and\ 1\ Black) = \frac{15}{32}$

(d) P(1st Red and 2nd Black)

This is calculated as:

$P(1st\ Red\ and\ 2nd\ Black) = [P(Red)\ and\ P(Black)]$

$P(1st\ Red\ and\ 2nd\ Black) = P(Red)\ *\ P(Black)$

$P(1st\ Red\ and\ 2nd\ Black) = \frac{n(Red)}{Total} *\ \frac{n(Black)}{Total}$

So, we have:

$P(1st\ Red\ and\ 2nd\ Black) = \frac{5}{8} *\frac{3}{8}$

$P(1st\ Red\ and\ 2nd\ Black) = \frac{15}{64}$

For probabilities without replacement

(a) P(2 Red)

This is calculated as:

$P(2\ Red) = P(Red)\ and\ P(Red)$

$P(2\ Red) = P(Red)\ *\ P(Red)$

So, we have:

$P(2\ Red) = \frac{n(Red)}{Total} \ *\ \frac{n(Red)-1}{Total-1}$

We subtracted 1 because the number of red balls (and the total) decreased by 1 after the first red ball is picked.

$P(2\ Red) = \frac{5}{8} * \frac{4}{7}$

$P(2\ Red) = \frac{5}{2} * \frac{1}{7}$

$P(2\ Red) = \frac{5}{14}$

(b) P(2 Black)

This is calculated as:

$P(2\ Black) = P(Black)\ and\ P(Black)$

$P(2\ Black) = P(Black)\ *\ P(Black)$

So, we have:

$P(2\ Black) = \frac{n(Black)}{Total}\ *\ \frac{n(Black)-1}{Total-1}$

We subtracted 1 because the number of black balls (and the total) decreased by 1 after the first black ball is picked.

$P(2\ Black) = \frac{3}{8}\ *\ \frac{2}{7}$

$P(2\ Black) = \frac{3}{4}\ *\ \frac{1}{7}$

$P(2\ Black) = \frac{3}{28}$

(c) P(1 Red and 1 Black)

This is calculated as:

$P(1\ Red\ and\ 1\ Black) = [P(Red)\ and\ P(Black)]\ or\ [P(Black)\ and\ P(Red)]$

$P(1\ Red\ and\ 1\ Black) = [P(Red)\ *\ P(Black)]\ +\ [P(Black)\ *\ P(Red)]$

$P(1\ Red\ and\ 1\ Black) = [\frac{n(Red)}{Total}\ *\ \frac{n(Black)}{Total-1}]\ +\ [\frac{n(Black)}{Total}\ *\ \frac{n(Red)}{Total-1}]$

So, we have:

$P(1\ Red\ and\ 1\ Black) = [\frac{5}{8} *\frac{3}{7}] + [\frac{3}{8} *\frac{5}{7}]$

$P(1\ Red\ and\ 1\ Black) = [\frac{15}{56} ] + [\frac{15}{56}]$

$P(1\ Red\ and\ 1\ Black) = \frac{30}{56}$

$P(1\ Red\ and\ 1\ Black) = \frac{15}{28}$

(d) P(1st Red and 2nd Black)

This is calculated as:

$P(1st\ Red\ and\ 2nd\ Black) = [P(Red)\ and\ P(Black)]$

$P(1st\ Red\ and\ 2nd\ Black) = P(Red)\ *\ P(Black)$

$P(1st\ Red\ and\ 2nd\ Black) = \frac{n(Red)}{Total} *\ \frac{n(Black)}{Total-1}$

So, we have:

$P(1st\ Red\ and\ 2nd\ Black) = \frac{5}{8} *\frac{3}{7}$

$P(1st\ Red\ and\ 2nd\ Black) = \frac{15}{56}$

7 0

3 years ago