1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Natali [406]
3 years ago
14

An atom is a small unit of matter. A small atom measures about 0.0000000001 meter. Write the decimal as a power of 10?

Mathematics
1 answer:
Lemur [1.5K]3 years ago
3 0
-0.01x10^9 just court how many zeros
You might be interested in
Do the ratios 4/5 and 12/20 form a proportion?
yKpoI14uk [10]

Answer:

No. They are not in proportion

Step-by-step explanation:

\sf \dfrac{4}{5} \ \text{is in the simplest form}

\sf \text{Simplest form of $\dfrac{12}{20}$} = \dfrac{12 \div 4}{20 \div 4}=\dfrac{3}{5}

So, they are not in proportion

7 0
3 years ago
Find the value of the expression below if I = 10.<br> z? – 2 (x + 5) =
12345 [234]

Answer:

10

Step-by-step explanation:

z−2(x+5)=

Use the distributive property to multiply −2 by x+5.

z−2x−10

3 0
3 years ago
A data set includes 103 body temperatures of healthy adult humans having a mean of 98.1 F and a standard deviation of 0.56 F. Co
Ira Lisetskai [31]

Answer:

The 99​% confidence interval estimate of the mean body temperature of all healthy humans is (97.955F, 98.245F).

98.6F is above the upper end of the interval, which means that the sample suggests that the mean body temperature could be lower than 98.6F.

Step-by-step explanation:

The first step is finding the confidence interval

The sample size is 103.

The first step to solve this problem is finding how many degrees of freedom there are, that is, the sample size subtracted by 1. So

df = 103-1 = 102

Then, we need to subtract one by the confidence level \alpha and divide by 2. So:

\frac{1-0.99}{2} = \frac{0.01}{2} = 0.005

Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 102 and 0.005 in the t-distribution table, we have T = 2.63.

Now, we need to find the standard deviation of the sample. That is:

s = \frac{0.56}{\sqrt{103}} = 0.055

Now, we multiply T and s

M = T*s = 2.63*0.055 = 0.145

For the lower end of the interval, we subtract the mean by M. So 98.1 - 0.145 = 97.955F.

For the upper end of the interval, we add the mean to M. So 98.1 + 0.145 = 98.245F.

The 99​% confidence interval estimate of the mean body temperature of all healthy humans is (97.955F, 98.245F).

98.6F is above the upper end of the interval, which means that the sample suggests that the mean body temperature could be lower than 98.6F.

5 0
3 years ago
I NEED THE ANSWER ASAP PLEASE
Natali [406]

Answer:

720$

Step-by-step explanation:

(1000+(230*24))-5800=720

8 0
3 years ago
Find the number of faces, edges, and vertices of this shape.​
sattari [20]
Faces: 7

Edges: 12

Vertices: 7
3 0
3 years ago
Read 2 more answers
Other questions:
  • Connie asked the students in a study group "how old are you?" each of them answered 12. is this a statistical question?
    12·1 answer
  • As a result of a cyclone, the price of bananas rose 85% to $9.99 per kg. How much did bananas cost before the cyclone?
    11·1 answer
  • Please help I dont understand<br><br> WILL GIVE BRAINLIEST
    8·1 answer
  • What is the value of sinC ?
    15·2 answers
  • A smoothie shop has 40 stores and 55% of the stores are in california the rest of the stores are in nevada.how many are in nevad
    13·1 answer
  • 18. Write down each pair of adjacent angles.​
    15·1 answer
  • Pls help me on this !!!!!!!!!!!
    11·1 answer
  • And adopt blinks about 450 times in 30 minutes a 12 year old blinks about 150 times in 15 minutes how many more times does an ad
    10·1 answer
  • There is 46% hard working Elfs in the North Pole. there is 300 elfs in north pole. How many elfs are hard working
    6·2 answers
  • Which output in this table is incorrect?
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!