All categories

2 years ago

In what way is ACCUPLACER an adaptive test?

Mathematics

2 answers:

notka56 [123]2 years ago

8 0

Answer:

The Accuplacer Math measures the student's ability to perform basic operations and solve problems involving mathematical concepts.

Step-by-step explanation:

The Accuplacer Math exam consists of three tests that measure the student's ability to perform basic operations and solve problems involving mathematical concepts.

Tests in Accuplacer Math exam:

1. The Arithmetic Test

2. The Quantitative Reasoning, Algebra, and Statistics Test

3. The Advanced Algebra and Functions Test.

Amanda [17]2 years ago

7 0

It presents students with questions based on their answers to previous questions

You might be interested in

Write the ratio of corresponding sides for the similar triangles and reduce the ratio to lowest terms.

serg [7]

Answer:

does it have a picture

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Help !!!!! What are the polar coordinates

OLga [1]

$\left(3,\,\frac{\big\pi}3\right)$

4 0

2 years ago

Round 2769.44 to the nearest tenths place

vlada-n [284]

It would be 2769.44 still since it would technically be 2769.440 if you were rounding to the TENTHS place. The Zero(0) doesn't change anything so the answer is

2769.44

4 0

3 years ago

PLS HELP THIS IS DUE TODAY!!!!!! NO LINKS, NO STEALING POINTS, AND ONLY CORRECT ANSWERS OR YOU WILL BE REPORTED!!!!!!!

Mnenie [13.5K]

Should be A I’ll show you what I did though it might not make sense because I kind of did it in a weird way but yeah that answer should be A hope this helps :)

3 0

1 year ago

Read 2 more answers

Suppose you have a light bulb that emits 50 watts of visible light. (Note: This is not the case for a standard 50-watt light bul

natali 33 [55]

Answer:

0.049168726 light-years

Step-by-step explanation:

The apparent brightness of a star is

$\bf B=\displaystyle\frac{L}{4\pi d^2}$

where

<em>L = luminosity of the star (related to the Sun) </em>

<em>d = distance in ly (light-years) </em>

The luminosity of Alpha Centauri A is 1.519 and its distance is 4.37 ly.

Hence the apparent brightness of Alpha Centauri A is

$\bf B=\displaystyle\frac{1.519}{4\pi (4.37)^2}=0.006329728$

According to the inverse square law for light intensity

$\bf \displaystyle\frac{I_1}{I_2}=\displaystyle\frac{d_2^2}{d_1^2}$

where

$\bf I_1=$ light intensity at distance $\bf d_1$

$\bf I_2=$ light intensity at distance $\bf d_2$

Let $\bf d_2$ be the distance we would have to place the 50-watt bulb, then replacing in the formula

$\bf \displaystyle\frac{0.006329728}{50}=\displaystyle\frac{d_2^2}{(4.37)^2}\Rightarrow\\\\\Rightarrow d_2^2=\displaystyle\frac{0.006329728*(4.37)^2}{50}\Rightarrow d_2^2=0.002417564\Rightarrow\\\\\Rightarrow d_2=\sqrt{0.002417564}=\boxed{0.049168726\;ly}$

Remark: It is worth noticing that Alpha Centauri A, though is the nearest star to the Sun, is not visible to the naked eye.

7 0

3 years ago