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pentagon [3]
2 years ago
8

Tim rolls a number cube labeled 1 through 6 and flips a coin. Which is the

Mathematics
1 answer:
ElenaW [278]2 years ago
4 0

There are 3 numbers on the cube greater than 3 ( 4, 5, & 6)

This means he has a 3/6 = 1/2 chance of doing that.

He has a 1/2 chance of getting tails on the coin.

The probability of doing both would be the probability of the cube roll x the probability of the coin toss:

1/2 x 1/2= 1/4

The answer is 1/4

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Johann wants to find a rectangular solar panel with an area of 128 cm. If the
Papessa [141]

Answer:

The length of rectangular panel is 6.4 cm.

Step-by-step explanation:

We are given the following information in the question:

Area of rectangular solar panel = 128 square cm

Width of panel = 20 cm

We have to find the length of the panel

Formula:

\text{Area of rectangular panel} = \text{Width}\times \text{Length}\\\Rightarrow 128 = 20\times \text{Length}\\\\\text{Length} = \displaystyle\frac{128}{20} = 6.4\text{ cm}

Hence, the length of rectangular panel is 6.4 cm.

This is an example of direct variation as greater would be length and breadth, the greater would be the area.

8 0
3 years ago
What is a complementary angle and a supplementary angle? Please write at least a paragraph about it.
Elis [28]

Answer:

Step-by-step explanation:

A comlemetery angle is a angel or angles that add up to 90 degrees. They make a right angle. A supplemetry angle is a agle or agles that add up to 180 degrees and make a strate line.

6 0
3 years ago
SCCoast, an Internet provider in the Southeast, developed the following frequency distribution on the age of Internet users. Fin
Tomtit [17]

Answer:

Mean: 40.17 years.

Standard deviation: 10.97 years.

Step-by-step explanation:

The frequency distribution is in the attached image.

We can calculate the mean adding the multiplication of midpoints of each class and frequency, and dividing by the sample size.

The midpoints of a class is calculated as the average of the bounds of the class.

Then, the mean can be written as:

E(X)=\dfrac{1}{N}\sum f_iX_i\\\\\\E(X)=\dfrac{1}{60}(3\cdot15+7\cdot25+18\cdot35+20\cdot45+12\cdot 55)=\dfrac{45+175+630+900+660}{60}\\\\\\E(X)=\dfrac{2,410}{60}=40.17

The standard deviation can be calculated as:

s=\sqrt{\dfrac{1}{N-1}\sum f_i(X_i-E(X))^2}\\\\\\s=\sqrt{\dfrac{1}{59}[3(15-40.17)^2+7(25-40.17)^2+18(35-40.17)^2+20(45-40.17)^2+12(55-40.17)^2]}

\\\\\\s=\sqrt{\dfrac{1}{59}( 1,900.59  + 1,610.90  + 481.12  + 466.58  + 2,639.15  )}\\\\\\s=\sqrt{\dfrac{ 7,098.33 }{59}}=\sqrt{120}=10.97

4 0
3 years ago
A group of randomly selected Apple Valley High School students were asked to pick their favorite gym class. The table below show
joja [24]

Answer:it is D)172 in the khan academy answers

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl; then a seco
kari74 [83]

Answer:

C. \frac{1}{18}

Step-by-step explanation:

Given: Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl then a second card is drawn.

To Find: If the cards are drawn at random and if the sum of the numbers on the cards is 8, what is the probability that one of the two cards drawn is numbered 5.

Solution:

Sample space for sum of cards when two cards are drawn at random is \{(1,1),(1,2),(1,3)......(6,6)\}

total number of possible cases =36

Sample space when sum of cards is 8 is \{(3,5),(5,3),(6,2),(2,6),(4,4)\}

Total number of possible cases =5

Sample space when one of the cards is 5 is \{(5,3),(3,5)\}

Total number of possible cases =2

Let A be the event that sum of cards is 8

p(\text{A}) =\frac{\text{total cases when sum of cards is 8}}{\text{all possible cases}}

p(\text{A})=\frac{5}{36}

Let B be the event when one of the two cards is 5

probability than one of two cards is 5 when sum of cards is 8

p(\frac{\text{B}}{\text{A}})=\frac{\text{total case when one of the number is 5}}{\text{total case when sum is 8}}

p(\frac{\text{B}}{\text{A}})=\frac{2}{5}

Now,

probability that sum of cards 8 is and one of cards is 5

p(\text{A and B}=p(\text{A})\times p(\frac{\text{B}}{\text{A}})

p(\text{A and B})=\frac{5}{36}\times\frac{2}{5}

p(\text{A and B})=\frac{1}{18}

if sum of cards is 8 then probability that one of the cards is 8 is \frac{1}{18}, option C is correct.

3 0
3 years ago
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