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mylen [45]
3 years ago
8

A spinner is divided into 8 equal sections, and each section contains a number from 1 to 8. What is the probability of the spinn

er landing on a number that is greater than 5?
StartFraction 1 over 13 EndFraction
StartFraction 1 over 8 EndFraction
StartFraction 3 over 13 EndFraction
StartFraction 3 over 8 EndFraction
Mathematics
2 answers:
lianna [129]3 years ago
8 0

Answer:

d

Step-by-step explanation:

Ad libitum [116K]3 years ago
5 0

Answer:

i think d he looks smart

Step-by-step explanation:

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if your bill at a restaurant is $56.43 and you want to leave 20% tipe,how much tipe should you leave​
morpeh [17]

\hspace{34em}|\\ \begin{array}{|c|ll} \cline{1-1} \textit{a\% of b}\\ \cline{1-1} \\ \left( \cfrac{a}{100} \right)\cdot b \\\\ \cline{1-1} \end{array}~\hspace{5em}\stackrel{\textit{20\% of 56.43}}{\left( \cfrac{20}{100} \right)56.43}\implies 11.286~\hfill \stackrel{about}{\approx 11.29}

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3 years ago
Marco has a collection of 437 bottles. Each month he buys 32 bottles .
Katarina [22]

Answer:

what is the question

Step-by-step explanation:

5 0
3 years ago
Show that the sum of the first 2n natural numbers is n(2n+1)
Leokris [45]

Answer:

This sum is the sum of an arithmetic sequence.  There is a formula for the sum of an arithmetic sequence which can be looked up or derived by a variety of means.

A nice approach for this sequence is the following.  Notice  that the sum of first and last number in the sequence is the same as the sum of the second and second last, and also the same as the sum of the third and third last, and so on.

There are n of these pairs.  So the desired sum is n x (first number + last number).   But the first number is 1 and the last on is 2n.   Thus the desired sum is n(1 + 2n).

Hope this helps!!

Mark Brainleast!!!!!!!!!!!

7 0
2 years ago
Read 2 more answers
How many 1/4 cup serving are in a 6 cup container
Ray Of Light [21]

Answer:

24

Step-by-step explanation:

6 ÷ 1/4 = 6 × 4 = 24

8 0
2 years ago
Read 2 more answers
I need answer Immediately pls!!!!!!
Illusion [34]

Given:

Total number of students = 27

Students who play basketball = 7

Student who play baseball = 18

Students who play neither sports = 7

To find:

The probability the student chosen at randomly from the class plays both basketball and base ball.

Solution:

Let the following events,

A : Student plays basketball

B : Student plays baseball

U : Union set or all students.

Then according to given information,

n(U)=27

n(A)=7

n(B)=18

n(A'\cap B')=7

We know that,

n(A\cup B)=n(U)-n(A'\cap B')

n(A\cup B)=27-7

n(A\cup B)=20

Now,

n(A\cup B)=n(A)+n(B)-n(A\cap B)

20=7+18-n(A\cap B)

n(A\cap B)=7+18-20

n(A\cap B)=25-20

n(A\cap B)=5

It means, the number of students who play both sports is 5.

The probability the student chosen at randomly from the class plays both basketball and base ball is

\text{Probability}=\dfrac{\text{Number of students who play both sports}}{\text{Total number of students}}

\text{Probability}=\dfrac{5}{27}

Therefore, the required probability is \dfrac{5}{27}.

3 0
3 years ago
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