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Brut [27]
2 years ago
14

Which answer shows 2.13786 times 10 Superscript 4 written in standard form? 0.000213786 2,137.86 21,378.6 2,137,860,000

Mathematics
2 answers:
Svetradugi [14.3K]2 years ago
6 0

Answer:

21378.6

Step-by-step explanation:

You move the decimal point four places towards the right.

podryga [215]2 years ago
4 0

Answer:

21378.6

Step-by-step explanation:

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Laura tiene en su cuenta ahorro del banco $23.500. Si paga el recibo de la luz que son $12.000
faltersainse [42]

Answer:

le queda -2

Step-by-step explanation:

porque 23.500 - 12.000 = 11.5 - 13.500 = -2

8 0
2 years ago
Two student council representatives are chosen at random from a group of 7 girls and 3 boys. Let G be the random variable denoti
Archy [21]

This question is incomplete, the complete question is;

Two student council representatives are chosen at random from a group of 7 girls and 3 boys. Let G be the random variable denoting the number of girls chosen.

a) What is E[G] or range of G

b) Give the distribution over the random variable G

Answer:

a) E[G] is [ 0, 1 , 2 ]

b)

the distribution over the random variable G.

x     P(x)

0     0.0667

1      0.4667

2     0.4667

Step-by-step explanation:

Given the data in the question;

Number to be chosen is 2

from 7 girls and 3 boys

a) range of G

number of girls can be; [ 0, 1 , 2 ]

Therefore, E[G] is [ 0, 1 , 2 ]

b) he distribution over the random variable G.

Distribution of the random variable G is Hypergeometric;

so

with P( G=g ) = P( getting g girls from 7 and 2-g boys from 3)

= ((^7C_g) × (^3C_{2-g)) / ^{10}C_2

now since, the range of G is [ 0, 1 , 2 ]

P( G=0 ) = ((^7C_0) × (^3C_{2)) / ^{10}C_2

= [(7!/(0!(7-0)!)) × (3!/(2!(3-2)!))] / (10!/(2!(10-2)!))

= [1 × 3] / 45 = 3/45 = 0.0667

P( G=1 ) = ((^7C_1) × (^3C_{1)) / ^{10}C_2

= [(7!/(1!(7-1)!)) × (3!/(1!(3-1)!))] / (10!/(2!(10-2)!))

= [ 7 × 3 ] / 45 = 21/45 = 0.4667

P( G=2 ) = ((^7C_2) × (^3C_{0)) / ^{10}C_2

= [(7!/(2!(7-2)!)) × (3!/(0!(3-0)!))] / (10!/(2!(10-2)!))

= [ 21 × 1 ] / 45 = 21/45 = 0.4667

Therefore, the distribution over the random variable G.

x     P(x)

0     0.0667

1      0.4667

2     0.4667

3 0
3 years ago
.00015 kilograms to grams​
Tpy6a [65]

Answer:

.15

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
The compressive strength of samples of cement can be modeled by a normal distribution with a mean of 6000 kg/cm2 and a variance
Anna [14]

Answer: the probability is 0.364

Step-by-step explanation:

Since the compressive strength of samples of cement can be modeled by a normal distribution, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = the compressive strength of samples of cement.

µ = mean compressive strength.

σ = standard deviation

From the information given,

µ = 6000 kg/cm²

Variance = 8100

σ = √variance = √8100 = 90

We want to find the probability that the compressive strength is between 5900 and 6000 kg/cm2. It is expressed as

P(5900 ≤ x ≤ 6000)

For x = 5900,

z = (5900 - 6000)/90 = - 1.11

Looking at the normal distribution table, the probability corresponding to the z score is 0.136

For x = 6000,

z = (6000 - 6000)/90 = 0

Looking at the normal distribution table, the probability corresponding to the z score is 0.5

P(5900 ≤ x ≤ 6000) = 0.5 - 0.136

P(5900 ≤ x ≤ 6000) = 0.364

6 0
2 years ago
Find the scale factors for the dilation of the small quadrilateral to the large
marysya [2.9K]
I think the correct answer to this question would be 2!
8 0
2 years ago
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