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Kaylis [27]
4 years ago
7

The made 9 pounds of fudge . The fudge was separated into 3/4 pound blocks . They sell each block for $6.50 .if they sell all th

e fudge ,how much money will they make

Mathematics
1 answer:
Flauer [41]4 years ago
3 0
Number 16, he put 1/2 of his money into his savings and took the rest to spend. So fraction of his amusement park money he spent on rides and popcorn is (1/5 + 3/4) x 1/2 = 19/40 <=> A

Each 3/4 pound block for $6.50. So each pound block will be $6.50 x 4/3 = 26/3. They have 9 pounds of fudge, money they will make is 26/3 x 9 = $78
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Fiona has $18 to spend. She spent $4.25, including tax, to buy a notebook. She needs to save $9.75, but she wants to buy a snack
chubhunter [2.5K]
18-4.25=13.75 the original minus what she already spent

13.75-9.75=4 that minus what she has to save

.50x=4 where x=the number of packages

divide both sides by .50
x=8 Fiona can buy 8 packages of crackers

Hope that helps.

6 0
3 years ago
Another name for the median in a set of data?. What is?
Nadya [2.5K]

Answer:

middle term

Step-by-step explanation:

1, 3, 3, 6, 7, 8, 9

In this example it would be 6

4 0
3 years ago
Prove or disprove (from i=0 to n) sum([2i]^4) &lt;= (4n)^4. If true use induction, else give the smallest value of n that it doe
ddd [48]

Answer:

The statement is true for every n between 0 and 77 and it is false for n\geq 78

Step-by-step explanation:

First, observe that, for n=0 and n=1 the statement is true:

For n=0: \sum^{n}_{i=0} (2i)^4=0 \leq 0=(4n)^4

For n=1: \sum^{n}_{i=0} (2i)^4=16 \leq 256=(4n)^4

From this point we will assume that n\geq 2

As we can see, \sum^{n}_{i=0} (2i)^4=\sum^{n}_{i=0} 16i^4=16\sum^{n}_{i=0} i^4 and (4n)^4=256n^4. Then,

\sum^{n}_{i=0} (2i)^4 \leq(4n)^4 \iff \sum^{n}_{i=0} i^4 \leq 16n^4

Now, we will use the formula for the sum of the first 4th powers:

\sum^{n}_{i=0} i^4=\frac{n^5}{5} +\frac{n^4}{2} +\frac{n^3}{3}-\frac{n}{30}=\frac{6n^5+15n^4+10n^3-n}{30}

Therefore:

\sum^{n}_{i=0} i^4 \leq 16n^4 \iff \frac{6n^5+15n^4+10n^3-n}{30} \leq 16n^4 \\\\ \iff 6n^5+10n^3-n \leq 465n^4 \iff 465n^4-6n^5-10n^3+n\geq 0

and, because n \geq 0,

465n^4-6n^5-10n^3+n\geq 0 \iff n(465n^3-6n^4-10n^2+1)\geq 0 \\\iff 465n^3-6n^4-10n^2+1\geq 0 \iff 465n^3-6n^4-10n^2\geq -1\\\iff n^2(465n-6n^2-10)\geq -1

Observe that, because n \geq 2 and is an integer,

n^2(465n-6n^2-10)\geq -1 \iff 465n-6n^2-10 \geq 0 \iff n(465-6n) \geq 10\\\iff 465-6n \geq 0 \iff n \leq \frac{465}{6}=\frac{155}{2}=77.5

In concusion, the statement is true if and only if n is a non negative integer such that n\leq 77

So, 78 is the smallest value of n that does not satisfy the inequality.

Note: If you compute  (4n)^4- \sum^{n}_{i=0} (2i)^4 for 77 and 78 you will obtain:

(4n)^4- \sum^{n}_{i=0} (2i)^4=53810064

(4n)^4- \sum^{n}_{i=0} (2i)^4=-61754992

7 0
4 years ago
To what decimal place should each answer be rounded? How many significant figures does the rounded answer have? A. 9 cm+2.8 cm=1
FromTheMoon [43]

Answer:

a) rounded to the ones place and 2 significant figures.

b) rounded to tenths place and 1 significant figure.

Step-by-step explanation:

Significant figures are the number of needed place values for accuracy. Round each number according to the least number of significant figures within the problem.

A. 9 cm+2.8 cm = 11.8 cm rounds to 12 which has 2 significant figures. Round to the ones place value since 9 has the least number of significant figures.

B. 0.135 atm+0.6 atm = 0.735 atm rounds to 0.7 atm which has 1 significant figure. Zero preceding the value is not considered significant. Round to the tenths place value since 0.6 has the least number of significant figures.

6 0
3 years ago
How do you figure out #9 and #10
Ann [662]
#9 the sum of two angles is 90 degrees, creating a right angle.

#10 the sum of the two angles is 180 because the form a supplementary angle
4 0
4 years ago
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