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sattari [20]
3 years ago
10

1 3 y 12 3 15 4 20??​

Mathematics
1 answer:
Lisa [10]3 years ago
5 0

•ω• Hewo fren!

☆☆●◉✿ Answer:✿◉●☆☆

Please make understandable questions.

☆☆●◉✿Step-by-step explanation:✿◉●☆☆

If you have a question and want a proper answer, you must make the question understandable first too. Made a mistake? Please tell me in the comments c:

HOPE I HELPED!  ∧∧

<h2>→⇒brainliest please? ∑(OΔO )♥♥︎</h2>

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At a new exhibit in the Museum of Science, people are asked to choose between 94 or 220 random draws from a machine. The machine
Tresset [83]

Answer:

0.0869 = 8.69% probability of getting more than 61% green balls.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

The machine is known to have 99 green balls and 78 red balls.

This means that p = \frac{99}{99+78} = 0.5593

Mean and standard deviation:

\mu = p = 0.5593

s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.5593*0.4407}{99+78}} = 0.0373

a. Calculate the probability of getting more than 61% green balls.

This is 1 subtracted by the pvalue of Z when X = 0.61. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{0.61 - 0.5593}{0.0373}

Z = 1.36

Z = 1.36 has a pvalue of 0.9131

1 - 0.9131 = 0.0869

0.0869 = 8.69% probability of getting more than 61% green balls.

8 0
3 years ago
I will give thanks and five stars to the person that helps me.
pentagon [3]

Answer:

Third option is correct!

Step-by-step explanation:

x+19<28

or, x<9

so

x = {3,4,5,6}

8 0
3 years ago
Multiply: (5x2y – 3xy2)(3xy2 + 5x2y)<br> Please help I need to get this one right!
Aliun [14]

Answer:

(1): "*-5x2y" was replaced by "*(-5x2y)".

(2): "x2"   was replaced by   "x^2".  1 more similar replacement(s).

STEP

1

:

Equation at the end of step 1

 (3x • (y2)) • (0 -  (5x2 • y))

STEP

2

:

Equation at the end of step

2

:

 3xy2 •  -5x2y

STEP

3

:

Multiplying exponential expressions :

3.1    x1 multiplied by x2 = x(1 + 2) = x3

Multiplying exponential expressions :

3.2    y2 multiplied by y1 = y(2 + 1) = y3

I believe the answer is   ( -3•5x3y3)

6 0
3 years ago
HELP ME PLEASE! AGH!
Vanyuwa [196]
I'm a third grader!!!!!!! Not a highschooler
8 0
3 years ago
Un elev a depus la banca suma de 250 lei. Știind ca banca oferă dobânda anuala de 1%,iar stația impozitează dobânda cu 16%,aflat
laila [671]

Răspuns:

292,50

Explicație pas cu pas:

Dobânda totală% = (1% + 16%) = 17% = 17/100 = 0,17

Suma principală depusă, P = 250

Suma totală, A după 1 an poate fi calculată astfel:

A = P (1 + rt)

Unde ; r = rata dobânzii; t = timp

A = 250 (1 + (0,17 * 1))

A = 250 (1,17)

A = 292,5

4 0
3 years ago
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