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3 years ago

A teacher gives a test with questions that fall into 3 levels of difficulty (easy, moderate, and hard).

1 answer:

4 0

When a teacher gives a test with questions that fall into 3 levels of difficulty (easy, moderate, and hard) The scoring method that would best allow her to identify the students who need help is:
B. weight each question by its difficulty: 1 for easy, 2 for moderate, and 3 for hard.

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How do i solve 6/13=3/c

Ostrovityanka [42]

You can do this by cross multiplying

3 0

3 years ago

if you travel to Jupiter, your weight would be 236.4% of your earth weight.How much would neil armstrong's space suit weigh on J

Margarita [4]

So,

This is simple multiplication once you've converted the percentage to a decimal.
236.4% = 2.364

Multiply 2.364 by 180.
2.364 * 180 = 425.52 lbs.

Neil Armstrong's space suit would weigh 425.52 lbs. on Jupiter.

4 0

3 years ago

How tall is the flagpole?<br> X<br> 6 ft<br> 15 ft<br> 4.5 ft

sweet-ann [11.9K]

Answer:

x = 20 ft

Step-by-step explanation:

We can use similar triangles and ratios to solve

4.5 ft 6 ft

--------- = ---------------

15 ft x ft

Using cross products

4.5x = 15*6

4.5x =90

Divide each side by 4.5

4.5x/4.5 = 90/4.5

x =20

3 0

3 years ago

Simplify. Evaluate the numerical bases. 3^2 b^2 c^-4 *3^-1 b^-5 c^7

photoshop1234 [79]

Hope I can be of assistance!

$3^2b^2c^{-4}\cdot \:3^{-1}b^{-5}c^7 \ \textgreater \ \mathrm{Apply\:exponent\:rule}:\ \:a^b\cdot \:a^c=a^{b+c}$
$3^{-1}\cdot \:3^2=\:3^{2-1}=\:3^1=\:3 \ \textgreater \ 3b^{-5}b^2c^{-4}c^7$

$\mathrm{Apply\:exponent\:rule}: \:a^b\cdot \:a^c=a^{b+c} \ \textgreater \ b^{-5}b^2=\:b^{2-5}=\:b^{-3}$
$3b^{-3}c^{-4}c^7$

$\mathrm{Apply\:exponent\:rule}: \:a^b\cdot \:a^c=a^{b+c} \ \textgreater \ c^{-4}c^7=\:c^{-4+7}=\:c^3$
$3b^{-3}c^3$

$\mathrm{Apply\:exponent\:rule}: \:a^{-b}=\frac{1}{a^b} \ \textgreater \ b^{-3}=\frac{1}{b^3} \ \textgreater \ 3\cdot \frac{1}{b^3}c^3$

$\mathrm{Multiply\:fractions}: \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c} \ \textgreater \ \frac{1\cdot \:3c^3}{b^3}$

Finally
$\mathrm{Apply\:rule}\:1\cdot \:a=a \ \textgreater \ \frac{3c^3}{b^3}$

Hope this helps!

8 0

3 years ago

Evaluate each expression using the values in the given table. x minus−3 minus−2 minus−1 0 1 2 3 f(x) minus−9 minus−7 minus−5 mi

Ymorist [56]

A) (f∘g)(-1) = f(g(-1)) = f(1) = -1
b) (g∘f)(1) = g(f(1)) = g(-1) = 1
c) (f∘f)(1) = f(f(1)) = f(-1) = -5

6 0

3 years ago