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2 years ago

111=13, 113=35, 115=57, 117=?

Mathematics

1 answer:

Elden [556K]2 years ago

3 0

In the pattern, the values are increasing by 22 each time so for the 117th term you take 57 + 22 = 79

So, 117= 79

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Required information Skip to question A die (six faces) has the number 1 painted on two of its faces, the number 2 painted on th

grigory [225]

Answer:

The change to the face 3 affects the value of P(Odd Number)

Step-by-step explanation:

Analysing the question one statement at a time.

Before the face with 3 is loaded to be twice likely to come up.

The sample space is:

$S = \{1,1,2,2,2,3\}$

And the probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{6}$

$P(1) = \frac{1}{3}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{6}$

$P(2) = \frac{1}{2}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{6}$

P(Odd Number) is then calculated as:

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{1}{3} + \frac{1}{6}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{6}$

$P(Odd\ Number) = \frac{3}{6}$

$P(Odd\ Number) = \frac{1}{2}$

After the face with 3 is loaded to be twice likely to come up.

The sample space becomes:

$S = \{1,1,2,2,2,3,3\}$

The probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

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

$P(2) = \frac{n(2)}{n(s)}$

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

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{7}$

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{2}{7} + \frac{1}{7}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{7}$

$P(Odd\ Number) = \frac{3}{7}$

Comparing P(Odd Number) before and after

$P(Odd\ Number) = \frac{1}{2}$ --- Before

$P(Odd\ Number) = \frac{3}{7}$ --- After

<em>We can conclude that the change to the face 3 affects the value of P(Odd Number)</em>

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2 years ago

PLZ HELP QUICK WILL MARK BRAINLIEST

slega [8]

For the Fist One it’s Pentagonal Pyramid

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3 years ago

Plz guys help i need asap<br> its part b im stuck

aleksklad [387]

Answer:

a.2nd quarter with 9 goals

b. 4.8 goals

c. 4 goals

Step-by-step explanation:

a. The mode is defined as the most appearing data point or the data point with the highest frequency..

From our data(for away goals):

1st quarter-2
2nd quarter-9
3rd quarter-7
4th quarter-4

Hence, the 2nd quarter has the mode for away goals with 9 goals.

b. Mean is defined as the average of a set of data points.

#We calculate the totals goals per quarter, sum over all quarters then divide by the number of games, 10:

$\bar x=\frac{1}{n}\sum{x_i}\\1^{st }_g=Away+Home=5+2=7\\\\2^{nd}_g=Away+Home=4+9=13\\\\3^{rd}_g=Away+Home=8+7=15\\\\4^{th}_g=Away+Home=9+4=13\\\\\bar x=\frac{1}{n}\sum{x_i}=\frac{1}{10}(7+13+15+13)=4.8$

Hence, the mean number of goals per quarter is 4.8 goals

c. To find the number of more home goals than away goals, we subtract from their summations as:

$g_m=\sum{g_h}-\sum{g_a}\\\\=(5+4+8+9)-(2+9+7+4)\\\\=26-22\\\\=4$

Hence, there are 4 more home goals than away goals.

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2 years ago

Nadia built a robot to filter air and water efficiently she expects the robot to filter more than 343 liter of air and water whi

crimeas [40]

Step-by-step explanation:

If she is able to achieve her aim, then the robot is very efficient, as it uses as low as 49 joules in filtering 343 litres of air and water.

It will now be a question of what time the robot is able to achieve this feat, if it is able to do it in a time that matches the quality of the work it does, then it will be reasonable to call the robot efficient, and fast too.

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3 years ago

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AysviL [449]

Answer:

96 seconds

Step-by-step explanation:

In order to find how long it will take to empty the bucket, you must multiply the number of gallons by the reciprocal of the fraction.

1. Reciprocal of $\frac{1}{3}$ is $\frac{3}{1}$ which is 3