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Alina [70]
3 years ago
14

Andrew has these marks: English 82% French 75% history 78% science 80% What mark will Andrew need in math if he wants his mean m

ark to be each percent? i)80% ii)81% iii)82%
Mathematics
1 answer:
Darina [25.2K]3 years ago
5 0
Sometimes, the strategy guess and check is the simplest. 82, 75, 78, and 80.
i) Let's try 80. All added together is 395 divided by 5 is 79%, which isn't enough. Now trying 85, the mean would be 80%.
ii) Let's try 90's mean. It's 81%.
iii) First, I'm going to try 95. The percentage mean is 82%
I didn't have to guess for ii) and iii) since I saw a pattern happening.
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H - 16 &lt; -24<br> solve this expression
Vikki [24]

Answer:

h < -8

Step-by-step explanation:

Add -16 to both sides to solve for h.

h - 16 + 16 < -24 + 16

h < -8

4 0
3 years ago
Your recipe for oatmeal raisin cookies calls for 3 pounds 3 ounces of rolled oats. Assuming you had enough of the other ingredie
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2 years ago
List 3 values that would make this inequality true y + 7 &gt; 18
blsea [12.9K]

Answer:It

It can be 12, 13, 14 and more up

Step-by-step explanation:

Is that because 12+7= 19 which is bigger than 18

3 0
3 years ago
If a is the average (arithmetic mean) of 4x and 7, b is the average of 5x and 6, and c is the average of 3x and 11, what is the
Debora [2.8K]

Answer:

2x+4

Step-by-step explanation:

Given that a is the average (arithmetic mean) of 4x and 7, b is the average of 5x and 6, and c is the average of 3x and 11

By definition of arithmetic mean,

we have a=\frac{4x+7}{2} \\b=\frac{5x+6}{2} \\c=\frac{3x+11}{2}

Let us calculate sum of a,b,c in terms of x

a+b+c = \frac{4x+7+5x+6+3x+11}{2} \\=6x+12

Average = sum/3 = \frac{6x+12}{3} =2x+4

7 0
3 years ago
If f (n)(0) = (n + 1)! for n = 0, 1, 2, , find the taylor series at a=0 for f.
Pie
Given that f^{(n)}(0)=(n+1)!, we have for f(x) the Taylor series expansion about 0 as

f(x)=\displaystyle\sum_{n=0}^\infty\frac{(n+1)!}{n!}x^n=\sum_{n=0}^\infty(n+1)x^n

Replace n+1 with n, so that the series is equivalent to

f(x)=\displaystyle\sum_{n=1}^\infty nx^{n-1}

and notice that

\displaystyle\frac{\mathrm d}{\mathrm dx}\sum_{n=0}^\infty x^n=\sum_{n=1}^\infty nx^{n-1}

Recall that for |x|, we have

\displaystyle\sum_{n=0}^\infty x^n=\frac1{1-x}

which means

f(x)=\displaystyle\sum_{n=1}^\infty nx^{n-1}=\frac{\mathrm d}{\mathrm dx}\frac1{1-x}
\implies f(x)=\dfrac1{(1-x)^2}
5 0
3 years ago
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