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postnew [5]
3 years ago
12

GRADE 7 MATH URGENT

Mathematics
2 answers:
crimeas [40]3 years ago
8 0

A. 8 pizza pies at 10 pc. per pie = 80 available pieces

B. Let X be the number of children. Number of pieces eaten = 5 per child = 5X.

c. 22 pieces are left over.

D. Combine line A and line C. Total number of pieces eaten = 80 original minus 22 left over. Result 58 pieces were eaten by X children: 58 = 5X.

E. Solve the equation 58 = 5X by dividing both sides by 5.

WHOOPS! Is this a trick question? Solving the above results in the answer that there are 11.6 children at the party. Hmmm. That 0.6 child is a question mark for me.

Looks like someone miscounted the pizza slices. Or omitted some pertinent information. But, I’m sure there is a logical answer.

WITCHER [35]3 years ago
7 0

Answer:

10 pizzas

Step-by-step explanation:

To find the answer for this question you need to multiply 3/8 by 25.

3/8 * 25 = 9.375

You can't buy .375 of a pizza so you round it up and get 10.

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mothers 8
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mothers 6
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4 0
3 years ago
Given the sequence in the table below, determine the sigma notation of the sum for term 4 through term 15.
Naya [18.7K]
It's a geometric sequence.

4,-12,36,... \\ \\
a_1=4 \\
r=\frac{a_2}{a_1}=\frac{-12}{4}=-3 \\ \\
a_n=a_1 \times r^{n-1} \\
a_n=4 \times (-3)^{n-1} \\
a_n=4 \times (-3)^{-1} \times (-3)^n \\
a_n=4 \times (-\frac{1}{3}) \times (-3)^n \\
a_n=-\frac{4}{3}(-3)^n

It's the sum for term 4 through term 15.

 \boxed{ \sum\limits_{n=4}^{15} (\frac{4}{3}(-3)^n)}
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Answer:

128

Step-by-step explanation:

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3 0
3 years ago
A positive integer from one to six is to be chosen by casting a die. Thus the elements c of the sample space C are 1, 2, 3, 4, 5
Alex_Xolod [135]

Answer with Step-by-step explanation:

We are given that six integers 1,2,3,4,5 and 6.

We are given that sample space

C={1,2,3,4,5,6}

Probability of each element=\frac{1}{6}

We have to find that P(C_1),P(C_2),P(C_1\cap C_2) \;and\; P(C_1\cup C_2)

Total number of elements=6

C_1={1,2,3,4}

Number of elements in C_1=4

P(E)=\frac{number\;of\;favorable \;cases}{Total;number \;of\;cases}

Using the formula

P(C_1)=\frac{4}{6}=\frac{2}{3}

C_2={3,4,5,6}

Number of elements in C_2=4

P(C_2)=\frac{4}{6}=\frac{2}{3}

C_1\cap C_2={3,4}

Number of elements in (C_1\cap C_2)=2

P(C_1\cap C_2)=\frac{2}{6}=\frac{1}{3}

C_1\cup C_2={1,2,3,4,5,6}

P(C_1\cup C_2)=\frac{6}{6}=1

4 0
3 years ago
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