All categories

3 years ago

Wendy has taken two tests with scores of 77 and 80. To get an 82

Mathematics

1 answer:

tiny-mole [99]3 years ago

3 0

Answer:

Step-by-step explanation:

Given that Wendy took 2 tests and scored, 77 and 80 respectively, let x represent the grade of Wendy's third test.

To get an average grade of 82, let's generate an equation to find x (Wendy's grade on her third test).

Average grade = sum of all 3 grades all over 3

$82 = \frac{77 + 80 + x}{3}$

$82 = \frac{157 + x}{3}$

Multiply 3 by both sides

$82*3 = \frac{157 + x}{3}*3$

$246 = 157 + x$

Subtract 157 from each side of the equation

$246 - 157 = 157 + x - 157$

$89 = x$

Wendy must score 89 on her third test to have an average of 82.

You might be interested in

HURRYYYYY

Troyanec [42]

Answer:

B is correct.

Step-by-step explanation:

Quadratic Expression (Standard Form) is:

$\displaystyle \large{a {x}^{2} + bx + c}$

To say it simply, the expression must have 2 as the highest degree.

This means that if there are any higher or lower degrees than 2 then they are not quadratic expression.

A choice is not quadratic expression because it has 1 as the highest degree.

B choice is correct because 2 is the highest degree.

C choice is wrong because 3 is the highest degree.

D choice is wrong because it is not a polynomial.

Therefore, B is correct.

7 0

2 years ago

What is 3/4 as a percentage

luda_lava [24]

Answer:

75%

Step-by-step explanation:

A percent is a value out of 100.

Change the fraction $\frac{3}{4}$ to an equivalent fraction with the denominator equal to 100.

$\frac{3}{4}$ × $\frac{25}{25}$ = $\frac{75}{100}$

This fraction represents a percent.

$\frac{75}{100}$ = 75%

5 0

2 years ago

8/4 divided by 100 equals what

dexar [7]

$\frac{8}{4}$ (8/4) divided by 100 equals .02

6 0

2 years ago

Read 2 more answers

A number x more than −13

Hunter-Best [27]

The expression is -13+x

7 0

3 years ago

An=an-1+6an-2 <br> n>2 a0=3 a1=6

Svetach [21]

$\begin{cases}a_0=3\\a_1=6\\a_n=a_{n-1}+6a_{n-2}&\text{for }n>2\end{cases}$

Let the generating function for $a_n$ be

$A(x)=\displaystyle\sum_{n\ge0}a_nx^n$

Multiplying both sides by $x^{n-2}$

$a_nx^{n-2}=a_{n-1}x^{n-2}+6a_{n-2}x^{n-2}$

Summing both sides over the non-negative integers greater than or equal to 2 gives

$\displaystyle\sum_{n\ge2}a_nx^{n-2}=\sum_{n\ge2}a_{n-1}x^{n-2}+6\sum_{n\ge2}a_{n-2}x^{n-2}$
$\displaystyle\frac1{x^2}\sum_{n\ge2}a_nx^n=\frac1x\sum_{n\ge1}a_nx^n+6\sum_{n\ge0}a_nx^n$
$\displaystyle\frac1{x^2}\left(\sum_{n\ge0}a_nx^n-a_0-a_1x\right)=\frac1x\left(\sum_{n\ge0}a_nx^n-a_0\right)+6\sum_{n\ge0}a_nx^n$
$\displaystyle\frac1{x^2}\left(A(x)-3-6x\right)=\frac1x\left(A(x)-3\right)+6A(x)$
$A(x)=\dfrac{3x+3}{1-x-6x^2}$
$A(x)=\dfrac3{5(1+2x)}+\dfrac{12}{5(1-3x)}$

For $|x|$ , the two series converge to

$\displaystyle A(x)=\frac35\sum_{n\ge0}(-2x)^n+\frac{12}5\sum_{n\ge0}(3x)^n$
$\implies A(x)=\displaystyle\sum_{n\ge0}\dfrac{3(-2)^n+12(3)^n}5x^n$
$a_n=\dfrac{3(-2)^n+12(3)^n}5=\dfrac{3(-2)^n+4(3)^{n+1}}5$

5 0

3 years ago