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Sergio039 [100]
3 years ago
15

2+2+3+4+4+5+5+8+9+5800= ??????????????? divided by 6 times 34 bb

Mathematics
1 answer:
sp2606 [1]3 years ago
4 0
33,104.6667 since you have to do step by step
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Where do I plot 1.25 and -1.25
slavikrds [6]

Answer:

I'd need to see more, but because they are the same number just positive and negative they'd be on opposite ends of zero

7 0
3 years ago
If cos(θ)=2853 with θin Q IV, what is sin(θ)?
marysya [2.9K]

Answer: \sin \theta=\frac{-45}{53}

Step-by-step explanation:

Since we have given that

\cos\theta=\frac{28}{53}

And we know that θ is in the Fourth Quadrant.

So, Except cosθ and sec θ, all trigonometric ratios will be negative.

As we know the "Trigonometric Identity":

\cos^2\theta+\sin^2\theta=1\\\\\sin \theta=\sqrt{1-\cos^2\theta}\\\\\sin \theta=\sqrt{1-(\frac{28}{53})^2}=\sqrt{\frac{53^2-28^2}{53^2}}\\\\\sin \theta=\sqrt{\frac{2025}{53^2}}\\\\\sin \theta=\frac{45}{53}

It must be negative due to its presence in Fourth quadrant.

Hence, \sin \theta=\frac{-45}{53}

7 0
3 years ago
A marble is selected at random from a jar containing 4 red marbles, 2 yellow marbles, and 3 green marbles.
s344n2d4d5 [400]

Answer:

4 / 9  probability of getting a red

Step-by-step explanation:

How many marbles are there total?

4 red + 2 yellow + 3 green = 9 marbles total

P( red marble) = 4 red / 9 total = 4/9

7 0
3 years ago
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What its two plus two
anastassius [24]

4 what do you think it is 1000000000

7 0
3 years ago
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An ordinary (fair) coin is tossed 3 times. Outcomes are thus triples of "heads" () and "tails" () which we write , , etc. For ea
boyakko [2]

Answer:

Some details are missing

Step-by-step explanation:

An ordinary (fair) coin is tossed 3 times. Outcomes are thus triples of "heads" (h) and "tails) (t) which we write hth, ttt, etc. For each outcome, let R be the random variable counting the number of heads in each outcome. For example, if the outcome is hht, then R(hht) = 2. Suppose that the random variable X is defined in terms of R as follows: X = 2R² - 6R - 1. The values of X are thus:

Outcome: || Value of X

tht || -5

thh || -5

hth || -5

htt || -5

hhh || -1

tth || -5

hht || -5

ttt || -1

Calculate the probability distribution function of X, i.e. the function Px (x). First, fill in the first row with the values of X. Then fill in the appropriate probabilities in the second row.

Solution

To calculate the probability distribution function of X.

We have to observe the total outcomes to check the number of "Heads (h) " in each outcome.

The first, fourth and, sixth outcome has 1 head (h)

The second, third and seventh outcome has 2 head (hh)

The fifth outcome has 3 head (hhh)

The eight outcome has 0 appearance of h

We then solve the probability of each occurrence

i.e. The probability of having h, hh, hhh and no occurrence of h

This will be represented as follows

P(h=0)

P(h=1)

P(h=2)

P(h=3)

In a coin, the probability of getting a head = ½ and the probability of getting a tail = ½ in 1 toss

Using the following formula

P(X=x) = nCr a^r * b ^ (n-r)

Where n represents total number of toss = 3

r represents number of occurrence

a represents getting a head = ½

b represents probability of getting a tail = ½

1. For h = 0

P(h=0) = 3C0 * ½^0 * ½³

P(h=0) = 1 * 1 * ⅛

P(h=0) = ⅛

2. For h = 1

P(h=1) = 3C1 * ½^1 * ½²

P(h=1) = 3 * ½ * ¼

P(h=1) = ⅜

3. P(h=2) = 3C2 * ½² * ½^1

P(h=2) = 3 * ¼ * ½

P(h=2) = ⅜

4.P(h=3) = 3C3 * ½³ * ½^0

P(h=0) = 1 * ⅛ * 1

P(h=0) = ⅛

It should be noted that when X is -5, h is either 1 or 2 and P(X) = ⅜

When X is -1, h is either 0 or 3 and P(X) = ⅛

The probability distribution function of X is as follows

Values of X || P(x)

-5 || ⅜

1 || ⅛

6 0
3 years ago
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