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3 years ago

ZDAC = ZBAD.

Mathematics

1 answer:

Troyanec [42]3 years ago

7 0

Answer:

Step-by-step explanation:

if 2+2 is 4 and 4+4 is 8 then the answer is c :)

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If 5(3x − 2) = 6, then x = a. 2/6 b. 5/6 c. 10/15 d. 16/15

kirill [66]

Answer:

Step-by-step explanation:

$5(3x-2)=6\\15x-10=6\\15x=16\\x=\frac{16}{15}$

5 0

3 years ago

Read 2 more answers

Help me plz i am very confused

Stells [14]

Well, if she wants to buy twice as many burgers as sodas, and her limit is $35, the inequality would have to be, how ever many burgers she can buy, while buying half that amount, but buying sodas instead..? I’m sorry if I’m confusing you more

8 0

3 years ago

Given that n^2/3 = 8 and n> 0

nalin [4]

Answer:

16 $\sqrt{2}$

Step-by-step explanation:

Using the rule of exponents/ radicals

$a^{\frac{m}{n} }$ = $\sqrt[n]{a^{m} }$ , then

$n^{\frac{2}{3} }$ = $\sqrt[3]{n^2}$ , then

$\sqrt[3]{n^2}$ = 8 ( cube both sides )

n² = 8³ ( take the square root of both sides )

n = $\sqrt{8^3}$

= $\sqrt{512}$

= $\sqrt{256(2)}$

= $\sqrt{256}$ × $\sqrt{2}$

= 16 $\sqrt{2}$

3 0

3 years ago

Growing from 1.4m to 1.6m in one year what’s the rate to change

jonny [76]

Answer:

1.6 - 1.4 = 0.2

0.2 / 1.4 = 0.14 = 14%

Step-by-step explanation:

6 0

3 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