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tensa zangetsu [6.8K]
3 years ago
12

PLZ ANSWER ASAP A bag contains 6 blue marbles, 10 red marbles, and 9 green marbles. If two marbles are drawn at random without r

eplacement, what is the probability that two red marbles are drawn?
Mathematics
2 answers:
LUCKY_DIMON [66]3 years ago
8 0

Answer:

10/25

10+9+6 equals 25 and there is 10 red marbles. I think this is the answer don't take my word for it.

svp [43]3 years ago
8 0

Answer:

I would determindedy say your answer is 3/25.

Step-by-step explanation:


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Can some answer pls.
kirill [66]

Answer:

E. (7,-2)

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
it is known that the population proton of utha residnet that are members of the church of jesus christ 0l6 suppose a random samp
Lady_Fox [76]

Answer:

0.0838 = 8.38% probability of obtaining a sample proportion less than 0.5.

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 establishes 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}}

Proportion of 0.6

This means that p = 0.6

Sample of 46

This means that n = 46

Mean and standard deviation:

\mu = p = 0.6

s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.6*0.4}{46}} = 0.0722

Probability of obtaining a sample proportion less than 0.5.

p-value of Z when X = 0.5. So

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

By the Central Limit Theorem

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

Z = \frac{0.5 - 0.6}{0.0722}

Z = -1.38

Z = -1.38 has a p-value of 0.0838

0.0838 = 8.38% probability of obtaining a sample proportion less than 0.5.

8 0
3 years ago
Find the surface area of the cylinder
STatiana [176]
<h2><u>Solution</u>:-</h2>

• Surface area of a cylinder = 2πr(r + h) sq. units.

In the above diagram,

Radius (r) of the cylinder = 13 cm.

Height (h) of the cylinder = 39 cm.

<h3>• Taking value of π = 3.14</h3>

Hence, Surface area of the cylinder = 2 × 3.14 × 13(13 + 39)

Surface area of the cylinder = (81.64 × 52) cm²

Surface area of the cylinder = 4245.28 cm²

The surface area of the cylinder is <u>4</u><u>2</u><u>4</u><u>5</u><u>.</u><u>2</u><u>8</u><u> </u><u>cm²</u>. [Answer]

6 0
2 years ago
Find how many six-digit numbers can be formed from the digits 2, 3, 4, 5, 6 and 7 (with repetitions), if:
Goshia [24]

Answer:

case 1 = 2592

case 2 =  729

case 1 + case 2 =  2916

(this is not a direct adition, because case 1 and case 2 have some shared elements)

Step-by-step explanation:

Case 1)

6 digits numbers that can be divided by 25.

For the first four positions, we can use any of the 6 given numbers.

For the last two positions, we have that the only numbers that can be divided by 25 are numbers that end in 25, 50, 75 or 100.

The only two that we can create with the numbers given are 25 and 75.

So for the fifth position we have 2 options, 2 or 7,

and for the last position we have only one option, 5.

Then the total number of combinations is:

C = 6*6*6*6*2*1 = 2592

case 2)

The even numbers are 2,4 and 6

the odd numbers are 3, 5 and 7.

For the even positions we can only use odd numbers, we have 3 even positions and 3 odd numbers, so the combinations are:

3*3*3

For the odd positions we can only use even numbers, we have 3 even numbers, so the number of combinations is:

3*3*3

we can put those two togheter and get that the total number of combinations is:

C = 3*3*3*3*3*3 = 3^6 = 729

If we want to calculate the combinations togheter, we need to discard the cases where we use 2 in the fourth position and 5 in the sixt position (because those numbers are already counted in case 1) so we have 2 numbers for the fifth position and 2 numbers for the sixt position

Then the number of combinations is

C = 3*3*3*3*2*2 = 324

Case 1 + case 2 = 324 + 2592 = 2916

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3 years ago
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3 years ago
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