Ask question

Ask question

All categories

3 years ago

13

Frazer scores 70% in a spelling test with 20 questions. How many did he get right? Show me your work.

1 answer:

ladessa [460]3 years ago

4 0

Fraser got 14 questions right

questions correct 70/5 = 14
—— ——
total questions 100/5 = 20

therefore he got 14 questions right

you divide 100 by 5 because percents are always out of 100 and that gets you 20 which is how many questions the spelling bee had

then you do the same to the top (divide by 5) and get 14

14/20 questions were answered right
70% of the questions were correct

You might be interested in

A student concluded that the solution to the equation sqrt2x+1 +3=0 is x=4. Do you agree? Why or why not?

Katyanochek1 [597]

Just put 4 as x in the equation and solve it. The new equation is:

sqrt(2*4)+1+3 = 0

Now we can solve it.

sqrt(2*4) becomes sqrt8:

sqrt8+1+3=0. -------> sqrt8+4=0

The sqrt of 8 is about 2.83.. so 2.83+4 is:

6.83

So we have 6.83 as the answer

but 6.83 does not equal 0 so x cannot equal 4

5 0

2 years ago

Natasha and her two dogs were walking on a perfectly straight road when her two dogs ran away from her in opposite directions. H

AlekseyPX

The answer to this question is B

5 0

3 years ago

Read 2 more answers

In tuv y is centroid if yw=5b and ty=4b+14 find tw

stellarik [79]

your right Step-by-step explanation:

3 0

2 years ago

Write two fractions that are equivalent to 5

Dima020 [189]

Answer:

5/1, 25/5

Step-by-step explanation:

8 0

3 years ago

The position of an object moving along an x axis is given by x = 3.24 t - 4.20 t2 + 1.07 t3, where x is in meters and t in secon

Pavel [41]

Answer and explanation:

Given : The position of an object moving along an x axis is given by $x=3.24t-4.20t^2+1.07t^3$ where x is in meters and t in seconds.

To find : The position of the object at the following values of t :

a) At t= 1 s

$x(t)=3.24t-4.20t^2+1.07t^3$

$x(1)=3.24(1)-4.20(1)^2+1.07(1)^3$

$x(1)=3.24-4.20+1.07$

$x(1)=0.11$

b) At t= 2 s

$x(t)=3.24t-4.20t^2+1.07t^3$

$x(2)=3.24(2)-4.20(2)^2+1.07(2)^3$

$x(2)=6.48-16.8+8.56$

$x(2)=-1.76$

c) At t= 3 s

$x(t)=3.24t-4.20t^2+1.07t^3$

$x(3)=3.24(3)-4.20(3)^2+1.07(3)^3$

$x(3)=9.72-37.8+28.89$

$x(3)=0.81$

d) At t= 4 s

$x(t)=3.24t-4.20t^2+1.07t^3$

$x(4)=3.24(4)-4.20(4)^2+1.07(4)^3$

$x(4)=12.96-67.2+68.48$

$x(4)=14.24$

(e) What is the object's displacement between t = 0 and t = 4 s?

At t=0, x(0)=0

At t=4, x(4)=14.24

The displacement is given by,

$\triangle x=x(4)-x(0)$

$\triangle x=14.24-0$

$\triangle x=14.24$

(f) What is its average velocity from t = 2 s to t = 4 s?

At t=2, x(2)=-1.76

At t=4, x(4)=14.24

The average velocity is given by,

$\triangle x=x(4)-x(2)$

$\triangle x=14.24-(-1.76)$

$\triangle x=14.24+1.76$

$\triangle x=16$

4 0

3 years ago