Answer:
YOU NEED TO SHOW THE DATA FOR THIS
Given:
Number of yellow balls = 10
Number of red balls = 5
Total number of outcomes = 10+5 = 15
Possible outcomes (for yellow balls) = 10
Probability of drawing yellow cards = 10/15
(Since there are 10 yellow balls)
= 2/3
The word "MORE" has 4 letters, therefore
Total number of outcomes = 4
Now, according to the situation:
M comes only once on the word MORE, so
Possible outcomes = 1
Probability = 1/4
[tex] \red{ \tiny A \scriptsize \: N \small \:S \large \: W \Large \:E \huge \: R}[/tex]
[tex] \rule{300pt}{0.1pt}[/tex]
sol. no:-3
To find mean of first six numbers.
so first odd no. is 1,3,5,7,9,11
[tex] \\ \tt \text{Mean} = \frac{1 + 3 + 5 + 7 + 9 + 11}{6} [/tex]
[tex] \\ \tt \qquad \quad = \frac{36}{6} [/tex]
[tex] \\ \boxed{ \red{\tt \text{Mean} = 6}}[/tex]
Solution no.:-4
Given : Temperature: [tex]\tt 55^{\circ} f , 47 \%, 80 f , 68^{\circ} F , 43^{\circ} f , 42^{\circ} F , 43^{\circ} F[/tex]
To find = mean temperature.
[tex] \\ \tt T_{m}=\frac{(55+47+80+68+43+42+43) {}^{ \circ} F }{7} [/tex]
[tex] \\ \tt=T_{m}=\frac{378}{7}=54^{\circ} F [/tex]
[tex] \pink{ \tt \[ \text { Hence mean temperature is } T_{m}=54^{\circ} F \text {. } \]}[/tex]
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Verified answer
Answer:
I)
YOU NEED TO SHOW THE DATA FOR THIS
II)
Given:
Number of yellow balls = 10
Number of red balls = 5
Total number of outcomes = 10+5 = 15
Possible outcomes (for yellow balls) = 10
Probability of drawing yellow cards = 10/15
(Since there are 10 yellow balls)
= 2/3
III)
The word "MORE" has 4 letters, therefore
Total number of outcomes = 4
Now, according to the situation:
M comes only once on the word MORE, so
Possible outcomes = 1
Probability = 1/4
IV)
[tex] \red{ \tiny A \scriptsize \: N \small \:S \large \: W \Large \:E \huge \: R}[/tex]
[tex] \rule{300pt}{0.1pt}[/tex]
sol. no:-3
To find mean of first six numbers.
so first odd no. is 1,3,5,7,9,11
[tex] \\ \tt \text{Mean} = \frac{1 + 3 + 5 + 7 + 9 + 11}{6} [/tex]
[tex] \\ \tt \qquad \quad = \frac{36}{6} [/tex]
[tex] \\ \boxed{ \red{\tt \text{Mean} = 6}}[/tex]
[tex] \rule{300pt}{0.1pt}[/tex]
Solution no.:-4
Given : Temperature: [tex]\tt 55^{\circ} f , 47 \%, 80 f , 68^{\circ} F , 43^{\circ} f , 42^{\circ} F , 43^{\circ} F[/tex]
To find = mean temperature.
[tex] \\ \tt T_{m}=\frac{(55+47+80+68+43+42+43) {}^{ \circ} F }{7} [/tex]
[tex] \\ \tt=T_{m}=\frac{378}{7}=54^{\circ} F [/tex]
[tex] \pink{ \tt \[ \text { Hence mean temperature is } T_{m}=54^{\circ} F \text {. } \]}[/tex]
[tex] \rule{300pt}{0.1pt}[/tex]