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KonstantinChe [14]
3 years ago
11

How many digits are there in the smallest sequence of repeating digits in the decimal equivalent.

Mathematics
1 answer:
Step2247 [10]3 years ago
7 0

the answer is 4

but I'm not for sure

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The formula for the area A of a triangle is A=(1/2)bh, where b is the length of the base and h is the height. Rearrange the form
zalisa [80]

Answer:

b=\dfrac{2A}{h}

Step-by-step explanation:

The formula for the area of a triangle is given by :

A=\dfrac{1}{2}bh ...(1)

Where b is the length of the base and h is the height.

We need to find the value of b from the above formula.

Cross multiplying equation (1) we get :

2A=bh

Now dividing both sides by h. So,

\dfrac{2A}{h}=\dfrac{bh}{h}\\\\b=\dfrac{2A}{h}

So, the correct option is (D) i.e. \dfrac{2A}{h}.

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Step-by-step explanation:

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3 years ago
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For f(x) = square root of x, estimate f(17) to the nearest integer.
Tomtit [17]

f(x) = sqrt(x)

f(17) = sqrt(17)

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3 years ago
The number of texts per day by students in a class is normally distributed with a 
kobusy [5.1K]

Answer:

1, 2, 6

Step-by-step explanation:

The z score shows by how many standard deviations the raw score is above or below the mean. The z score is given by:

z=\frac{x-\mu}{\sigma} \\\\where\ x=raw\ score,\mu=mean, \ \sigma=standard\ deviation

Given that mean (μ) = 130 texts, standard deviation (σ) = 20 texts

1) For x < 90:

z=\frac{x-\mu}{\sigma} \\\\z=\frac{90-130}{20} =-2

From the normal distribution table, P(x < 90) = P(z < -2) = 0.0228 = 2.28%

Option 1 is correct

2) For x > 130:

z=\frac{x-\mu}{\sigma} \\\\z=\frac{130-130}{20} =0

From the normal distribution table, P(x > 130) = P(z > 0) = 1 - P(z < 0) = 1 - 0.5 = 50%

Option 2 is correct

3) For x > 190:

z=\frac{x-\mu}{\sigma} \\\\z=\frac{190-130}{20} =3

From the normal distribution table, P(x > 3) = P(z > 3) = 1 - P(z < 3) = 1 - 0.9987 = 0.0013 = 0.13%

Option 3 is incorrect

4)  For x < 130:

z=\frac{x-\mu}{\sigma} \\\\z=\frac{130-130}{20} =0

For x > 100:

z=\frac{x-\mu}{\sigma} \\\\z=\frac{100-130}{20} =-1.5

From the normal table, P(100 < x < 130) = P(-1.5 < z < 0) = P(z < 0) - P(z < 1.5) = 0.5 - 0.0668 = 0.9332 = 93.32%

Option 4 is incorrect

5)  For x = 130:

z=\frac{x-\mu}{\sigma} \\\\z=\frac{130-130}{20} =0

Option 5 is incorrect

6)  For x = 130:

z=\frac{x-\mu}{\sigma} \\\\z=\frac{160-130}{20} =1.5

Since 1.5 is between 1 and 2, option 6 is correct

5 0
2 years ago
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