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3 years ago

11

Final grades are often computed using weighted averages. For example, Fran's teacher uses the following: homework 10%, quizzes 2

0%, and tests 70%.
Fran has an average homework score of 92. Her quiz score average is 68, and her test score average is 81. What final grade can Fran expect?
A) 73.5
B) 75.5
C) 78
D) 79.5

1 answer:

STALIN [3.7K]3 years ago

8 0

The final grade of fran can be calculated by multiplying the corresponding percentage of his average homework, quizzes and tests, then sum it up and it is now the final grade of fran.

Final grade = (0.10 x 92) + ( 0.20 x 68) + (0.70 x 81)

<span>Final grade = 79.5 </span>

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A two digit number has a tens digit and a ones digit.
Let's say x = tens digit and y = ones digit

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The number is: 10x + y
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Then you can use substitution or elimination to combine and solve.
I'll use elimination, multiply the entire top equation by -3 and add the equations together. y will cancel out

-3x - 3y = -15
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3 years ago

What is 9.6+5.42-3.2 squared

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139.7124

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2 years ago

Read 2 more answers

Let Q(x, y) be the predicate "If x < y then x2 < y2," with domain for both x and y being R, the set of all real numbers.

Levart [38]

Answer:

a) Q(-2,1) is false

b) Q(-5,2) is false

c)Q(3,8) is true

d)Q(9,10) is true

Step-by-step explanation:

Given data is $Q(x,y)$ is predicate that $x$ then $x^{2}$ . where $x,y$ are rational numbers.

a)

when $x=-2, y=1$

Here $-2$ that is $x$ satisfied. Then

$(-2)^{2}$

$4$ this is wrong. since $4>1$

That is $x^{2}$ $>y^{2}$ Thus $Q(x,y)$ $=Q(-2,1)$ is false.

b)

Assume $Q(x,y)=Q(-5,2)$ .

That is $x=-5, y=2$

Here $-5$ that is $x$ this condition is satisfied.

Then

$(-5)^{2}$

$25$ this is not true. since $25>4$ .

This is similar to the truth value of part (a).

Since in both $x$ satisfied and $x^{2} >y^{2}$ for both the points.

c)

if $Q(x,y)=Q(3,8)$ that is $x=3$ and $y=8$

Here $3$ this satisfies the condition $x$ .

Then $3^{2}$

$9$ This also satisfies the condition $x^{2}$ .

Hence $Q(3,8)$ exists and it is true.

d)

Assume $Q(x,y)=Q(9,10)$

Here $9$ satisfies the condition $x$

Then $9^{2}$

$81$ satisfies the condition $x^{2}$ .

Thus, $Q(9,10)$ point exists and it is true. This satisfies the same values as in part (c)

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