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3 years ago

Answer the following questions CORRECTLY I will know if this is wrong. I WILL REPORT ANY INCORRECT ANSWERS!

Mathematics

1 answer:

Anna71 [15]3 years ago

8 0

Answer by JKismyhusbandbae: B) –10.6

Work/Explanation: Since the sequence slowly gets smaller, it is likely that each term is something added to (subtracted from) the previous term. To get from –7.9 to –8.8, it appears that the first has had –0.9 added to it. Continue adding –0.9 to get that the fourth term is –10.6.

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SIMPLE HELP HELP HELP PLEASE ONLY 1 ANSWER NONE OF THE ABOVE OR ALL IS NOT THE ANSWER SO PLEASE DO IT RIGHT THIS TIME WILL GIVE

Eva8 [605]

Answer:

The first option is the correct!

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3 years ago

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How do you simplify the value of (28x^4)÷(12x^3)?

d1i1m1o1n [39]

$(28x^4) \div (12x^3) = \dfrac{28x^4}{12x^3} = \dfrac{4 \times 7x^4}{4 \times 3 x^3} = \dfrac{7}{3} x^{4-3} = \dfrac{7}{3} x$

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3 years ago

The Patronete Winery's tastiest wine must have a 12% alcohol content. How many gallons of wine with a 9% alcohol content must be

erik [133]

Answer: 3000 gallon

Step-by-step explanation:

Here, The total quantity of 15% alcohol content = 3000 gallon

And, in which quantity of alcohol = 15% of 3000 gallon = 450 gallon.

Let the total quantity of 9% alcohol content = x

In which quantity of alcohol = 9% of x gallon= 9x/100 gallon

Now, according to the question, The wine with a 9% alcohol content will be mixed with 3,000 gallons of wine with a 15% alcohol content in order to achieve the desired 12% alcohol content.

Thus, the total quantity of mixture of 12% alcohol content = The total quantity of 15% alcohol content + total quantity of 9% alcohol content

= 3000 + x

In which quantity of alcohol = 12 % of ( 3000+x) = (360 + 12x/100) gallon ---(1)

But, the quantity of alcohol in 12 % of alcohol content = quantity of alcohol in 15% of alcohol content + quantity of alcohol in 9% of alcohol content

The quantity of alcohol in 12 % of alcohol content = (450 + 9x/100) gallon ---(2)

On equating equation (1) and (2)

We get, 360 + 12x/100 = 450 + 9x/100

⇒ 3x/100 = 90

⇒ x = 3000

Therefore, The total quantity of the mixture of 9% alcohol content = 3000

6 0

3 years ago

Due to a manufacturing error, two cans of regular soda were accidentally filled with diet soda and placed into a 18-pack. Suppos

crimeas [40]

Answer:

a) There is a 1.21% probability that both contain diet soda.

b) There is a 79.21% probability that both contain diet soda.

c) $P(X = 2)$ is unusual, $P(X = 0)$ is not unusual

d) There is a 19.58% probability that exactly one is diet and exactly one is regular.

Step-by-step explanation:

There are only two possible outcomes. Either the can has diet soda, or it hasn't. So we use the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And $\pi$ is the probability of X happening.

A number of sucesses $x$ is considered unusually low if $P(X \leq x) \leq 0.05$ and unusually high if $P(X \geq x) \geq 0.05$

In this problem, we have that:

Two cans are randomly chosen, so $n = 2$

Two out of 18 cans are filled with diet coke, so $\pi = \frac{2}{18} = 0.11$

a) Determine the probability that both contain diet soda. P(both diet soda)

That is P(X = 2).

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 2) = C_{2,2}(0.11)^{2}(0.89)^{0} = 0.0121$

There is a 1.21% probability that both contain diet soda.

b)Determine the probability that both contain regular soda. P(both regular)

That is P(X = 0).

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 0) = C_{2,0}(0.11)^{0}(0.89)^{2} = 0.7921$

There is a 79.21% probability that both contain diet soda.

c) Would this be unusual?

We have that $P(X = 2)$ is unusual, since $P(X \geq 2) = P(X = 2) = 0.0121 \leq 0.05$

For P(X = 0), it is not unusually high nor unusually low.

d) Determine the probability that exactly one is diet and exactly one is regular. P(one diet and one regular)

That is P(X = 1).

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 1) = C_{2,1}(0.11)^{1}(0.89)^{1} = 0.1958$

There is a 19.58% probability that exactly one is diet and exactly one is regular.

8 0

3 years ago