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3 years ago

How do I solve the problem?

Mathematics

2 answers:

Irina-Kira [14]3 years ago

5 0

Sal: 25 +5x = 240

Divide 5 from both sides so it would be 5+ x=240/5
Subtract 5 from both sides x=48-5
X= 43

Elena: 15 + 5x=300

Divide 5 from both sides so it would be 3 +x=300/5
Subtract 15 from both sides x=60-3
X=57

So Sal jogged for 43 minutes. While Elena jogged for 57 minutes

tekilochka [14]3 years ago

3 0

X will be 57 after analyzing this problem!

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Answer:

Step-by-step explanation:

Method 1:

$p\%=\dfrac{p}{100}\\\\30\%=\dfrac{30}{100}=\dfrac{3}{10}=0.3\\\\30\%\ of\ 70\ is\\\\0.3\cdot70=21$

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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