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
alexdok [17]
3 years ago
7

7÷2761 long division

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

Answer:

0.0025

Step-by-step explanation:

You might be interested in
The price of a Tesla Model X was bought for \$80,000 and has depreciated 10.2% yearly. Find the price of the car 3 years later
ella [17]

Answer:

A is the correct answer ...

6 0
2 years ago
Hannah purchased three items at the store for $7.05, $8.12, and $3.97. She paid for
vichka [17]

Answer:

She should receive $30.86

Step-by-step explanation:

$7.05 + $8.12 + $3.97 =19.14

50 - 19.14 = 30.86

6 0
3 years ago
A student answers 80% of the questions on the math exam correctly, if she answers 40 questions correctly. How many questions are
mafiozo [28]
Hope this helps! ask if you need clarification.

6 0
3 years ago
The five values for a data set are:
dem82 [27]

Answer:

The box-and-whisker plot for the given question is as shown at the attached figure.

Minimum = 11

Lower quartile = Q1 = 12

Median = Q2 = 23.5

Upper quartile = Q3 = 27

Maximum = 33

<u>So, according to the given data and the figure:</u>

  • The box will go from <u>12 to 27</u>                ⇒ Q1 to Q3
  • A line dividing the box will go at <u>23.5</u>   ⇒ Q2
  • The left whisker will go from <u>11 to 12</u>     ⇒ Minimum to Q1
  • The right whisker will go from to <u>27 to 33</u>  ⇒ Q2 to Maximum

5 0
3 years ago
Read 2 more answers
A 95% confidence interval was computed using a sample of 16 lithium batteries, which had a sample mean life of 645 hours. The co
polet [3.4K]

Answer:

Step-by-step explanation:

Hello!

The mean life of 16 lithium batteries was estimated with a 95% CI:

(628.5, 661.5) hours

Assuming that the variable "X: Duration time (life) of a lithium battery(hours)" has a normal distribution and the statistic used to estimate the population mean was s Student's t, the formula for the interval is:

[X[bar]±t_{n-1;1-\alpha /2}* \frac{S}{\sqrt{n} }]

The amplitude of the interval is calculated as:

a= Upper bond - Lower bond

a= [X[bar]+t_{n-1;1-\alpha /2}* \frac{S}{\sqrt{n} }] -[X[bar]-t_{n-1;1-\alpha /2}* \frac{S}{\sqrt{n} }]

and the semiamplitude (d) is half the amplitude

d=(Upper bond - Lower bond)/2

d=([X[bar]+t_{n-1;1-\alpha /2}* \frac{S}{\sqrt{n} }] -[X[bar]-t_{n-1;1-\alpha /2}* \frac{S}{\sqrt{n} }] )/2

d= t_{n-1;1-\alpha /2}* \frac{S}{\sqrt{n} }

The sample mean marks where the center of the calculated interval will be. The terms of the formula that affect the width or amplitude of the interval is the value of the statistic, the sample standard deviation and the sample size.

Using the semiamplitude of the interval I'll analyze each one of the posibilities to see wich one will result in an increase of its amplitude.

Original interval:

Amplitude: a= 661.5 - 628.5= 33

semiamplitude d=a/2= 33/2= 16.5

1) Having a sample with a larger standard deviation.

The standard deviation has a direct relationship with the semiamplitude of the interval, if you increase the standard deviation, it will increase the semiamplitude of the CI

↑d= t_{n-1;1-\alpha /2} * ↑S/√n

2) Using a 99% confidence level instead of 95%.

d= t_{n_1;1-\alpha /2} * S/√n

Increasing the confidence level increases the value of t you will use for the interval and therefore increases the semiamplitude:

95% ⇒ t_{15;0.975}= 2.131

99% ⇒ t_{15;0.995}= 2.947

The confidence level and the semiamplitude have a direct relationship:

↑d= ↑t_{n_1;1-\alpha /2} * S/√n

3) Removing an outlier from the data.

Removing one outlier has two different effects:

1) the sample size is reduced in one (from 16 batteries to 15 batteries)

2) especially if the outlier is far away from the rest of the sample, the standard deviation will decrease when you take it out.

In this particular case, the modification of the standard deviation will have a higher impact in the semiamplitude of the interval than the modification of the sample size (just one unit change is negligible)

↓d= t_{n_1;1-\alpha /2} * ↓S/√n

Since the standard deviation and the semiamplitude have a direct relationship, decreasing S will cause d to decrease.

4) Using a 90% confidence level instead of 95%.

↓d= ↓t_{n_1;1-\alpha /2} * S/√n

Using a lower confidence level will decrease the value of t used to calculate the interval and thus decrease the semiamplitude.

5) Testing 10 batteries instead of 16. and 6) Testing 24 batteries instead of 16.

The sample size has an indirect relationship with the semiamplitude if the interval, meaning that if you increase n, the semiamplitude will decrease but if you decrease n then the semiamplitude will increase:

From 16 batteries to 10 batteries: ↑d= t_{n_1;1-\alpha /2} * S/√↓n

From 16 batteries to 24 batteries: ↓d= t_{n_1;1-\alpha /2} * S/√↑n

I hope this helps!

4 0
3 years ago
Other questions:
  • How do i graph this
    12·1 answer
  • 1) 7+a=15<br><br><br><br> 2) 23=d+44<br><br><br><br> 3) 28=n-14<br><br><br><br> 4) T-22=-31
    12·2 answers
  • Consider the statement | x − 4 | &lt; 0.5 |x-4|&lt;0.5. What is the meaning of this statement? Select all that apply. a. All val
    7·1 answer
  • Ebonae fish tank holds 40 gallons.how many quarts does the fish tank hold?
    8·1 answer
  • More math help please im lazy ._.
    14·1 answer
  • Is this relation a function?
    7·1 answer
  • The width of a rectangle is 2 inches smaller than its length. The perimeter is 24 inches. Find the length and width of the recta
    6·1 answer
  • How is the graphed system of linear equations classified? Drag and drop words into the boxes to correctly complete the statement
    9·2 answers
  • Pls answer ........... ​
    11·2 answers
  • (Please someone help me!) (No links!)
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!