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olchik [2.2K]
3 years ago
13

Does anyone know 81,452 divided by 93

Mathematics
2 answers:
arsen [322]3 years ago
6 0

Answer:

875

Step-by-step explanation:

Vesnalui [34]3 years ago
4 0
The answer should be 875.82
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Ok, i need to determine if it can be written as a fraction. Then state the reason.
zhannawk [14.2K]

Answer:

a) YES. Repeating decimal

b) YES. Repeating decimal

c) NO. Decimal neither terminates nor repeats

d) YES. Terminating decimal

Step-by-step explanation:

Any rational number can be written as a fraction, that is, a division of integers.

a) The number \overline{0.54} has a repetitive period of 54. It can be expressed as a fraction because its decimal are repeating forever

b) The number 0.16666... (expressed as a repeating 6) can also be converted to a fraction because it's a repeating decimal

c) The number 0.5473... cannot be expressed as a fraction because the ellipsis (...) means the decimals continue with no limit and no pattern can be found (no repetitive decimal)

d) The number 0.378 can be easily expressed as a fraction because it has a finite number of decimals (no ellipsis) or a terminating decimal

6 0
3 years ago
Find the smallest sample size n that will guarantee at least a 90% chance of the sample mean income being within $500 of the pop
Andreyy89

Answer:

The smallest sample size n that will guarantee at least a 90% chance of the sample mean income being within $500 of the population mean income is 48.

Step-by-step explanation:

The complete question is:

The mean salary of people living in a certain city is $37,500 with a standard deviation of $2,103. A sample of n people will be selected at random from those living in the city. Find the smallest sample size n that will guarantee at least a 90% chance of the sample mean income being within $500 of the population mean income. Round your answer up to the next largest whole number.

Solution:

The (1 - <em>α</em>)% confidence interval for population mean is:

CI=\bar x\pm z_{\alpha/2}\cdot\frac{\sigma}{\sqrt{n}}

The margin of error for this interval is:

MOE=z_{\alpha/2}\cdot\frac{\sigma}{\sqrt{n}}

The critical value of <em>z</em> for 90% confidence level is:

<em>z</em> = 1.645

Compute the required sample size as follows:

MOE=z_{\alpha/2}\cdot\frac{\sigma}{\sqrt{n}}

      n=[\frac{z_{\alpha/2}\cdot\sigma}{MOE}]^{2}\\\\=[\frac{1.645\times 2103}{500}]^{2}\\\\=47.8707620769\\\\\approx 48

Thus, the smallest sample size n that will guarantee at least a 90% chance of the sample mean income being within $500 of the population mean income is 48.

3 0
3 years ago
In a class of 35 students 1/5
dangina [55]

Answer:

7 play cricket

14 play football

21 play volleyball

Step-by-step explanation:

4 0
3 years ago
Anyone please answer me ,<br> Please solve this question
Mamont248 [21]

Answer:

67

Step-by-step explanation:

Multi-Inscribed Quadrilateral: 15

Bananas: 4

Clock: 3

Therefore, we can substitute for the last equation the values, so it would become:

3 + 4 + 4 x 15 = ?

First, we solve the multiplication according to PEMDAS.

4 x 15 = 60

Then, we can add from left to right.

3 + 4 + 60 = ?

3 + 4  = 7

7 + 60 = 67

Therefore, our answer would be 67.

3 0
3 years ago
Read 2 more answers
. In a class of 40 students, 8 are in the drama club and 12 are in the art club. If a student is selected at random, what is the
Levart [38]

Answer:

so the probability of selected student is in the drama club is 1/5

Step-by-step explanation:

total students = 40

drama club= 8

art club=  12

we have to find probability of drama club

P (drama) = favorable outcome / total outcome

                 =   8 / 40

                = 1/ 5

so the probability of selected student is in the drama club is 1/5

4 0
4 years ago
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