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kifflom [539]
3 years ago
6

Select the equivalent expression. Thanks :)

Mathematics
1 answer:
Umnica [9.8K]3 years ago
6 0

Answer:

b

Step-by-step explanation:

(y^3 * 2^5)^2

3x2=6

5x2=10

y6x2^10

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A multiple choice test has four possible answers for each question. Let X be the number of correct responses in a test of 100 qu
dimulka [17.4K]

Answer:

a. 4.33.

Step-by-step explanation:

If there is only one correct answer out of four alternatives, the expected proportion of correct answers is p =0.25

Sample size (n) = 100 questions

The standard deviation of a proportion 'p' is given by:

s=\sqrt{\frac{p*(1-p)}{n} }

Applying the given data:

s=\sqrt{\frac{0.25*(1-0.25)}{100}}\\s=0.0433

If X is the number of correct responses in 100 guesses, the standard deviation of X is:

\sigma = n*s=100*0.0433\\\sigma = 4.33

The standard deviation of X is 4.33 questions.

7 0
3 years ago
D/d{cosec^-1(1+x²/2x)} is equal to​
SIZIF [17.4K]

Step-by-step explanation:

\large\underline{\sf{Solution-}}

\rm :\longmapsto\:\dfrac{d}{dx} {cosec}^{ - 1} \bigg( \dfrac{1 +  {x}^{2} }{2x} \bigg)

Let assume that

\rm :\longmapsto\:y =  {cosec}^{ - 1} \bigg( \dfrac{1 +  {x}^{2} }{2x} \bigg)

We know,

\boxed{\tt{  {cosec}^{ - 1}x =  {sin}^{ - 1}\bigg( \dfrac{1}{x} \bigg)}}

So, using this, we get

\rm :\longmapsto\:y = sin^{ - 1} \bigg( \dfrac{2x}{1 +  {x}^{2} } \bigg)

Now, we use Method of Substitution, So we substitute

\red{\rm :\longmapsto\:x = tanz \: \rm\implies \:z =  {tan}^{ - 1}x}

So, above expression can be rewritten as

\rm :\longmapsto\:y = sin^{ - 1} \bigg( \dfrac{2tanz}{1 +  {tan}^{2} z} \bigg)

\rm :\longmapsto\:y = sin^{ - 1} \bigg( sin2z \bigg)

\rm\implies \:y = 2z

\bf\implies \:y = 2 {tan}^{ - 1}x

So,

\bf\implies \: {cosec}^{ - 1}\bigg( \dfrac{1 +  {x}^{2} }{2x} \bigg) = 2 {tan}^{ - 1}x

Thus,

\rm :\longmapsto\:\dfrac{d}{dx} {cosec}^{ - 1} \bigg( \dfrac{1 +  {x}^{2} }{2x} \bigg)

\rm \:  =  \: \dfrac{d}{dx}(2 {tan}^{ - 1}x)

\rm \:  =  \: 2 \: \dfrac{d}{dx}( {tan}^{ - 1}x)

\rm \:  =  \: 2 \times \dfrac{1}{1 +  {x}^{2} }

\rm \:  =  \: \dfrac{2}{1 +  {x}^{2} }

<u>Hence, </u>

\purple{\rm :\longmapsto\:\boxed{\tt{ \dfrac{d}{dx} {cosec}^{ - 1} \bigg( \dfrac{1 +  {x}^{2} }{2x} \bigg) =  \frac{2}{1 +  {x}^{2} }}}}

<u>Hence, Option (d) is </u><u>correct.</u>

6 0
3 years ago
What is the ratio of 20cm to 1m​
Julli [10]

9514 1404 393

Answer:

  1 to 5

Step-by-step explanation:

20 cm to 1 m = 20 cm to 100 cm = 20 to 100 = 1 to 5

3 0
3 years ago
Find the solution set for x2 + 3x – 40 = 0.
Temka [501]
I am going to assume that x2 = x squared
So: X^2 + 3x -40 
= (X+8)(X-5)=0

X+8 = 0
X= - 8

The other solution is X -5 = 0
X = 5

Therefore the solution set is {-8, 5}
8 0
4 years ago
In order to prepare for your summer bash, you go to the supermarket to buy hamburgers and
timofeeve [1]

The complete question is:

In order to prepare for your summer bash, you go to the supermarket to buy hamburgers and chicken. Hamburgers cost $2 per pound and chicken costs $3 per pound. You have no more than $30 to spend.  You expect to purchase at least 3 pounds of hamburgers.  

  • Write a system of inequalities to represent this situation.
  • Graph the system of inequalities on the grid.
  • Give three possible combinations for buying hamburgers and chicken for your summer bash.

Justify your answers

Answer:

A) System of inequalities:

  • 2x + 3y ≤ 30
  • x ≥ 3
  • x ≥ 0
  • y ≥ 0

B) Graph: see the picture attached

C) Three possible combinations:

  • 3 pounds of hamburgers and 8 pounds of chicken
  • 3 pounds of hamburgers and 0 pounds of chicken
  • 15 pounds of hamburgers and 0 pounds of chicken

Explanation:

<u>A) Write the system of inequalities to represent this situaction.</u>

<u>1. Variables:</u>

  • x: number of pound of hamburgers
  • y: number of pound of chicken

<u>2. Costs:</u>

  • x pounds of hamburgers at $2 per pound: 2x
  • y: pounds of chicken at $3 per pound: 3y

  • total cost: 2x + 3y

<u>3. First constraint:</u>

  • You have no more than $ 30 to spend: means that the cost of what you buy can be at most (less than or equal to) $ 30.

  • 2x + 3y ≤ 30

<u>4. Second constraint:</u>

  • You expect to purchase at least 3 pounds of hamburgers: means that the number of pounds of hamburgers may be greater than or equal to 3.

  • x ≥ 3

<u>5. Additional constraints:</u>

  • Both, x and y cannot be negative: x, y ≥ 0

<u>6. System of equations:</u>

  • 2x + 3y ≤ 30
  • x ≥ 3
  • x ≥ 0
  • y ≥ 0

<u>B) Graph </u>

You have to graph all the constraints in a x-y coordinate system.

<u>1. To graph 2x + 3y ≤ 30 graph the line 2x + 3y = 30</u>

  • Choose the y-intercept and x-intercept.
  • x = 0 ⇒ 3y = 30 ⇒ y = 10 ⇒ point (0, 10)
  • y = 0 ⇒ 2x =30 ⇒ x = 15 ⇒ point (15, 0)
  • With two points you can draw the line
  • Clear y: y ≤ 10 - 2x/3. Since, the symbol is ≤ you shade the region below the line, and the line is included, so you draw it as as solid line.

<u>2. To graph x ≥ 3 just draw the vertical line x = 3 and shade the region to the right of it. The points of the line are included (solid line).</u>

<u>3. The constraints x ≥ 0 and y ≥ 0 </u>mean that the region is restricted to the first quadrant (including the positive axis).

<u>4. The feasible solutions</u> are the set of points inside the common regions (intersection).

With all that information the graph is the one attached. The feasible solutions is the region defined by the triangle with vertices (3,8), (3,0), and (15,0).

<u>C) Give 3 possible combinations.</u>

You can pick any three points inside the region, as long as the coordinates are integer numbers. For instance the 3 vertices are solutions:

  • (3, 8) ⇒ 3 pounds of hamburgers and 8 pounds of chicken
  • (3,0) ⇒ 3 pounds of hamburgers and 0 pounds of chicken
  • (15,0) ⇒ 15 pounds of hamburgers and 0 pounds of chicken

You could also prove that (5,4), 5 pounds of hamburgers and 4 pounds of chicken meet, the inequalities.

This is how you prove it:

  • 5 ≥ 0
  • 5 ≥ 3
  • 4 ≥ 0
  • 2(5) + 3(4) = 10 + 12 = 22 ≤ 30

And you can do the same for any pairs to verify whether they are solution or not.

7 0
4 years ago
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