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2 years ago

Solve for x: 30 > 6x

Mathematics

2 answers:

Eddi Din [679]2 years ago

7 0

Answer:

If it has the little line under the > sign then x = 5

Step-by-step explanation:

Pani-rosa [81]2 years ago

4 0

here is the answer to solve for x x<5

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This is for 100 Which equation represents the relationship between the x-and y-values shown in the table

kicyunya [14]

Answer:

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Step-by-step explanation:

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2 years ago

Sandra is showing her work in simplifying (3.3 − 4.2) − 2.8 + 5.7. Identify any error in her work or reasoning. Step 1: (3.3 − 4

julsineya [31]

The answer is correct
but I think in step 2 that should read the Associative property

8 0

3 years ago

Read 2 more answers

Which is the more reasonable measurement of the distance between the tracks on a railroad:

nataly862011 [7]

If you're comparing 1.435 * 10^(-3) mm to 1.435 * 10^3 mm, then the more reasonable measurement is 1.435 * 10^3 mm since,

1.435 * 10^3 mm = 1435 mm = 1.435 meters

1 meter is about 3.28 feet approximately

So 1.435 meters is slightly larger (about 4.708 feet) which is a fairly reasonable distance between the tracks on a railroad.

-------------------

Something like 1.435 * 10^(-3) mm is equal to 0.001435 mm, which is far less than 1 mm.

6 0

3 years ago

Please help and thanks!~<br> ~~~~~~~~~~~~~~~~~~~~~~~<br> 20 points!~

andriy [413]

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8 0

3 years ago

4. In a sample of 1000 people, 130 can wiggle their ears. Two unrelated people are

Elina [12.6K]

Using the hypergeometric distribution, it is found that there is a 0.7568 = 75.68% probability that neither can wiggle his or her ears.

The people are chosen from the sample without replacement, which is why the <u>hypergeometric distribution</u> is used to solve this question.

Hypergeometric distribution:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}$

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

The parameters are:

x is the number of successes.
N is the size of the population.
n is the size of the sample.
k is the total number of desired outcomes.

In this problem:

1000 people means that $N = 1000$
130 can wiggle their ears, thus $k = 130$
Two are selected, thus $n = 2$ .

The probability that neither can wiggle his or her ears is P(X = 0), thus:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}$

$P(X = x) = h(0,1000,2,130) = \frac{C_{130,0}C_{870,2}}{C_{1000,2}} = 0.7568$

0.7568 = 75.68% probability that neither can wiggle his or her ears.

A similar problem is given at brainly.com/question/24826394

5 0

2 years ago