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3 years ago

How to write .0000321 in scientific notation?

1 answer:

5 0

Scientific notation that's involved with numbers less than one, should always be counted from left to right. Count starting from the decimal until the first number you "jump over". What are you counting exactly?- the number of digits between the decimal and the first 'number' other than zero.

0.0000321 can be also written as 3.21*10^-5 (or 10 with a negative 5 exponent)
Hope this helps.

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The coordinates of the preimage of polygon ABCD and the image ofA'B'C'D' are given below. What rule describes the transformation

I am Lyosha [343]

The coordinates of polygon ABCD are:

A(4,3) B(4,0) C(-1, 1) D(-1, 4)

and

The coordinates of the transformed polygon A'B'C'D' are:

A'(1, 1) B'(1, -2) C'(-4,-1) D'(-4, 2)

To understand the transformed rule, just remove the coordinates of the actual polygon from the transformed polygon.

A' - A = (1, 1) - (4, 3) = (1 - 4, 1 - 3) = (-3, -2)

Similarly,

B' - B = (1, -2) - (4, 0) = (1 - 4, -2 - 0) = (-3, 2)

The same result will be obtained from other points (C and D). Therefore, the first option (-3, 2) is correct.

3 0

1 year ago

A random sample of 36 students at a community college showed an average age of 25 years. Assume the ages of all students at the

Pavel [41]

Answer:

98% confidence interval for the average age of all students is [24.302 , 25.698]

Step-by-step explanation:

We are given that a random sample of 36 students at a community college showed an average age of 25 years.

Also, assuming that the ages of all students at the college are normally distributed with a standard deviation of 1.8 years.

So, the pivotal quantity for 98% confidence interval for the average age is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample average age = 25 years

$\sigma$ = population standard deviation = 1.8 years

n = sample of students = 36

$\mu$ = population average age

So, 98% confidence interval for the average age, $\mu$ is ;

P(-2.3263 < N(0,1) < 2.3263) = 0.98

P(-2.3263 < $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < 2.3263) = 0.98

P( $-2.3263 \times {\frac{\sigma}{\sqrt{n} }$ < ${\bar X - \mu}$ < $2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ) = 0.98

P( $\bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} }$ < $\mu$ < $\bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ) = 0.98

98% confidence interval for $\mu$ = [ $\bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} }$ , $\bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ]

= [ $25 - 2.3263 \times {\frac{1.8}{\sqrt{36} }$ , $25 + 2.3263 \times {\frac{1.8}{\sqrt{36} }$ ]

= [24.302 , 25.698]

Therefore, 98% confidence interval for the average age of all students at this college is [24.302 , 25.698].

8 0

3 years ago

Julio practiced his trumpet on each 6 days during a week. What is the mean of minutes Julio practiced over these 6 days.

Novay_Z [31]

21+30+39+43+58+67 = 258.
258/6 = 43

43 minutes

4 0

3 years ago

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For the right triangle shown, the lengths of two sides are given. find the third side. leave your answer in simplified, radical

Citrus2011 [14]

There is no diagram

6 0

2 years ago

You plan to build a house that is 1 ½ times as long as it is wide. You want the land around the house to be 20 feet wider than t

IgorLugansk [536]

The total area with land based on the information given is 3x² + 60x.

Let the width = x

Let the length = (1.5 × x) = 1.5x

Area = 1.5x × x = 1.5x²

The area along with land:

Width = x + 20

Length = 3 × x = 3x

The total area with land:

= Length × Width

= 3x(x + 20)

= 3x² + 60x.

Learn more about length on:

brainly.com/question/2217700

#SPJ1

7 0

1 year ago