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2 years ago

I need to know how to do number 20

Mathematics

1 answer:

zimovet [89]2 years ago

6 0

$(x^4y^3)^2=(x^4)^2(y^3)^2=x^{4\cdot2}y^{3\cdot2}=x^8y^6\\\\Used:\\\\(a\cdot b)^n=a^n\cdot b^n\\\\(a^n)^m=a^{n\cdot m}$

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24 points<br><br> Which diagram BEST models 4x2/3?

Hoochie [10]

D) because the final answer should be 8/3. See the attachment.

4 0

2 years ago

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Geometry math question please help

Rom4ik [11]

Since BD joins the midpoints of two sides of a triangle, it is half the length of the third side, FE.

BD = (1/2)FE = (1/2)(23.5) = 11.75

Answer: B. 11.75

8 0

3 years ago

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If a student randomly guesses at 20 multiple-choice questions, find the probability that the student gets exactly four correct.

ad-work [718]

Answer:

Probability = 0.190

Step-by-step explanation:

Given:

Number of questions, 'n' = 20

Each question has four choices.

Let event of choosing correct answer be success and its probability be represented by 'p'.

Therefore, event of choosing wrong answer is failure and its probability be represented by 'q'.

Now, probability of success is choosing one correct answer out of 4 options. So, $p=\frac{1}{4}=0.25$

Now, 'p' and 'q' are complements of each other. Therefore,

$q=1-p=1-0.25=0.75$

Now, we need 4 successes. Therefore, $x=4$

Using Bernoulli's theorem to find 'x' successes from 'n' questions, we get:

$P(x)=_{x}^{n}\textrm{C}p^xq^{n-x}$

Plug in 4 for 'x', 20 for 'n', 0.25 for 'p' and 0.75 for 'q'. Solve.

$P(4)=_{4}^{20}\textrm{C}(0.25)^4(0.75)^{20-4}\\P(4)=_{4}^{20}\textrm{C}(0.25)^4(0.75)^{16}\\P(4)=4845\times (0.25)^4\times (0.65)^{16}\\\\P(4)=0.1896\approx0.190$

Therefore, the probability of getting 4 correct answers out of 20 questions is 0.190.

6 0

2 years ago

During a recent eruption, the volcano spewed out copious amount of ash. One small piece of ash was ejected from the volcano with

nlexa [21]

For question number 1:The plot H = H(t) is the parabola and it reaches its maximum in the moment when exactly at midpoint between the roots t = 0 and t = 23. At that moment t = 23/2 or 11.5 seconds.
For question number 2:To find the maximal height, just simply substitute t = 11.5 into the quadratic equation. The answer would be 22.9.
For question number 3:H(t) = 0, or, which is the same as -16t^2 + 368t = 0.Factor the left side to get -16*t*(t - 23) = 0.t = 0, relates to the very start of the process, when the ash started its way up.The other root is t = 23 seconds, and it is precisely the time moment when the bit of ash will go back to the ground.

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3 years ago

Graph the circle x^2+y^2+2x+4y-44=0

Natasha_Volkova [10]

Answer:

See explanation

Step-by-step explanation:

First, convert this to standard form. By completing the square, you can do the following:

$(x^2+2x+1)-1+(y^2+4y+4)-4-44=0$