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2 months ago

Which expression is equivalent to the given polynomial expression?

Mathematics

1 answer:

Zina [12.3K]2 months ago

6 0

C. -31m⁴n - 8m²Step-by-step explanation:Given:(9mn - 19m⁴n) - (8m² + 12m⁴n + 9mn)Required:Identify an equivalent expression for itSolution:Distributing the negative sign across the parentheses results in:9mn - 19m⁴n - 8m² - 12m⁴n - 9mnNext, we combine like terms:9mn - 9mn - 19m⁴n - 12m⁴n - 8m²This simplifies to -31m⁴n - 8m²Thus, -31m⁴n - 8m² is the equivalent expression for (9mn - 19m⁴n) - (8m² + 12m⁴n + 9mn).

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A reporter determines a baseball player's batting average, which is a ratio of number of hits to the number of times at bats. Th

Leona [12618]

x:200

x/200 =.2121212121...

200(x/200) = 200(.2121212121...)

x = 42.4242...

The expectation is that a player would achieve 42 hits over 200 at-bats.

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2 months ago

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a multiple choice test contains 50 questions with 4 answer choices. what is the probability of correctly answering 15 questions

Zina [12379]

The likelihood of answering each question correctly stands at 1/4, given that there is 1 correct answer out of 4 choices. Therefore, the probability of answering 15 questions correctly is computed as (1/4)^15 = 1/1073741824.

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1 month ago

a falling object travels a distance given by the formula d=5t + 16t^2 ft, where t is measured in seconds. how long will it take

zzz [12365]

It is provided that

$d=5t+16t^{2}$

We need to determine t when d equals 74

Thus, $74=5t+16t^{2}$ or

$16t^{2}+5t-74 = 0$

$16t^{2}+37t-32t-74 = 0$

t(16t + 37) - 2(16t + 37) = 0

(t - 2)(16t + 37) = 0

t - 2 = 0 or 16t + 37 = 0

t cannot be negative.

Thus, t equals 2

Therefore, it will require 2 seconds for the object to cover 74 feet.

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3 months ago

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You’re working in a pharmacy, and are making a table to help with dosage amounts for a certain drug. The recommended dosage is 4

lawyer [12517]

Answer: Children who weigh 10 pounds require 12 teaspoons, those at 15 pounds need 18 teaspoons, and for 30 pounds, it is necessary to have 36 teaspoons.

Step-by-step explanation:

We have a recommended dose of 40 mg per 2.2 pounds of weight, given in three doses throughout the day every 8 hours.

Thus, for a single pound, the daily required dosage is calculated as $\dfrac{40}{2.2}=18.18\ ml$ .

Each individual dose corresponds to $\dfrac{18.18}{3}=6.06\ ml$ of the drug.

This means for one pound, the dosage needed per dose is 6.06 ml.

Thus, for 10 pounds, the calculation is 10 x 6.06 ml = 60.6 ml

To convert ml to teaspoons, knowing that 1 teaspoon equals 5 ml, we find: (60.6 ml ÷ 5) = 12.12 ≈ 12 teaspoons.

For a weight of 15 pounds, the drug requirement is 15 x 6.06 ml = 90.9 ml.

This translates to (90.9 ml ÷ 5) = 18.18 ≈ 18 teaspoons.

For the weight of 30 pounds, it is calculated as 30 x 6.06 ml = 181.8 ml.

Converting this gives us (181.8 ml ÷ 5) = 36.36 ≈ 36 teaspoons.

Therefore, children weighing 10 pounds need 12 teaspoons, those at 15 pounds require 18 teaspoons, and those at 30 pounds need 36 teaspoons.

7 0

3 months ago

In Gallup's Annual Consumption Habits Poll, telephone interviews were conducted for a random sample of 1,014 adults aged 18 and

babunello [11817]

Answer:

(b)E(x)=1.3087

(c)Variance of x =1.7119

(d)E(y)=2.0447

Step-by-step explanation:

The random variable x refers to the average number of cups of coffee consumed daily.

The total number of respondents equals 1014

(a) Probability distribution for x.

$\left|\begin{array}{c|cccccc}x&0&1&2&3&4\\\\P(x)&\dfrac{365}{1014}&\dfrac{264}{1014}&\dfrac{193}{1014}&\dfrac{91}{1014}&\dfrac{101}{1014} \end{array}\right|$

(b) Expected value for x

$E(x)=\left(0\times\dfrac{365}{1014}\left)+\left(1\times\dfrac{264}{1014}\left)+\left(2\times\dfrac{193}{1014}\left)+\left(3\times\dfrac{91}{1014}\left)+\left(4\times\dfrac{101}{1014}\right)$

E(x)=1.3087

Variance $=\sum (x-\mu)^2P(x)$

$\left|\begin{array}{c|cccccc}x&0&1&2&3&4\\(x-\mu)^2&1.7167&0.0953&0.4779&2.8605&7.2431\\\\P(x)&\dfrac{365}{1014}&\dfrac{264}{1014}&\dfrac{193}{1014}&\dfrac{91}{1014}&\dfrac{101}{1014} \\\\(x-\mu)^2P(x)&0.6179&0.0248&0.0910&0.2567&0.7215\end{array}\right|$

$Variance, \sum (x-\mu)^2P(x)=1.7119$

(d)

$\left|\begin{array}{c|cccccc}y&1&2&3&4\\\\P(y)&\dfrac{264}{649}&\dfrac{193}{649}&\dfrac{91}{649}&\dfrac{101}{649} \end{array}\right|$

$E(y)=\left(1\times\dfrac{264}{649}\left)+\left(2\times\dfrac{193}{649}\left)+\left(3\times\dfrac{91}{649}\left)+\left(4\times\dfrac{101}{649}\right)$

E(y)=2.0447

The expected value for y surpasses that of x.

6 0

2 months ago