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
Galina-37 [17]
3 years ago
7

The 4 - pack is priced at £2.04

Mathematics
1 answer:
Archy [21]3 years ago
7 0
The 4 Pack is the better value ;)
You might be interested in
(8.6x107)-(9.1x10-8)simplify
natima [27]

Answer:

  85,999,999.999 999 909

Step-by-step explanation:

The expression represents the difference of a relatively large number and one that is relatively small. That difference is approximately the value of the large number. The exact value requires 17 digits for its proper expression. Most calculators and spreadsheets cannot display this many digits.

<h3>Standard form</h3>

The numbers in standard form are ...

  86,000,000 = 8.6×10^7

  0.000000091 = 9.1×10^-8

<h3>Difference</h3>

Their difference is ...

  86,000,000 -0.000000091 = 85,999,999.999 999 909

In scientific notation, this is ...

  8.599 999 999 999 990 9×10^7

4 0
2 years ago
Can someone pls answer this​
SVEN [57.7K]

Answer:

1/6

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
The author drilled a hole in a die and filled it with a lead​ weight, then proceeded to roll it 199 times. Here are the observed
Anton [14]

Answer with explanation:

An Unbiased Dice is Rolled 199 times.

Frequency of outcomes 1,2,3,4,5,6 are=28​, 29​, 47​, 40​, 22​, 33.

Probability of an Event

      =\frac{\text{Total favorable Outcome}}{\text{Total Possible Outcome}}\\\\P(1)=\frac{28}{199}\\\\P(2)=\frac{29}{199}\\\\P(3)=\frac{47}{199}\\\\P(4)=\frac{40}{199}\\\\P(5)=\frac{22}{199}\\\\P(6)=\frac{33}{199}\\\\\text{Dice is fair}\\\\P(1,2,3,4,5,6}=\frac{33}{199}

→→→To check whether the result are significant or not , we will calculate standard error(e) and then z value

1.

(e_{1})^2=(P_{1})^2+(P'_{1})^2\\\\(e_{1})^2=[\frac{28}{199}]^2+[\frac{33}{199}]^2\\\\(e_{1})^2=\frac{1873}{39601}\\\\(e_{1})^2=0.0472967\\\\e_{1}=0.217478\\\\z_{1}=\frac{P'_{1}-P_{1}}{e_{1}}\\\\z_{1}=\frac{\frac{33}{199}-\frac{28}{199}}{0.217478}\\\\z_{1}=\frac{5}{43.27}\\\\z_{1}=0.12

→→If the value of z is between 2 and 3 , then the result will be significant at 5% level of Significance.Here value of z is very less, so the result is not significant.

2.

(e_{2})^2=(P_{2})^2+(P'_{2})^2\\\\(e_{2})^2=[\frac{29}{199}]^2+[\frac{33}{199}]^2\\\\(e_{2})^2=\frac{1930}{39601}\\\\(e_{2})^2=0.04873\\\\e_{2}=0.2207\\\\z_{2}=\frac{P'_{2}-P_{2}}{e_{2}}\\\\z_{2}=\frac{\frac{33}{199}-\frac{29}{199}}{0.2207}\\\\z_{2}=\frac{4}{43.9193}\\\\z_{2}=0.0911

Result is not significant.

3.

(e_{3})^2=(P_{3})^2+(P'_{3})^2\\\\(e_{3})^2=[\frac{47}{199}]^2+[\frac{33}{199}]^2\\\\(e_{3})^2=\frac{3298}{39601}\\\\(e_{3})^2=0.08328\\\\e_{3}=0.2885\\\\z_{3}=\frac{P_{3}-P'_{3}}{e_{3}}\\\\z_{3}=\frac{\frac{47}{199}-\frac{33}{199}}{0.2885}\\\\z_{3}=\frac{14}{57.4279}\\\\z_{3}=0.24378

Result is not significant.

4.

(e_{4})^2=(P_{4})^2+(P'_{4})^2\\\\(e_{4})^2=[\frac{40}{199}]^2+[\frac{33}{199}]^2\\\\(e_{4})^2=\frac{3298}{39601}\\\\(e_{4})^2=0.06790\\\\e_{4}=0.2605\\\\z_{4}=\frac{P_{4}-P'_{4}}{e_{4}}\\\\z_{4}=\frac{\frac{40}{199}-\frac{33}{199}}{0.2605}\\\\z_{4}=\frac{7}{51.8555}\\\\z_{4}=0.1349

Result is not significant.

5.

(e_{5})^2=(P_{5})^2+(P'_{5})^2\\\\(e_{5})^2=[\frac{22}{199}]^2+[\frac{33}{199}]^2\\\\(e_{5})^2=\frac{1573}{39601}\\\\(e_{5})^2=0.03972\\\\e_{5}=0.1993\\\\z_{5}=\frac{P'_{5}-P_{5}}{e_{5}}\\\\z_{5}=\frac{\frac{33}{199}-\frac{22}{199}}{0.1993}\\\\z_{5}=\frac{11}{39.6610}\\\\z_{5}=0.2773

Result is not significant.

6.

(e_{6})^2=(P_{6})^2+(P'_{6})^2\\\\(e_{6})^2=[\frac{33}{199}]^2+[\frac{33}{199}]^2\\\\(e_{6})^2=\frac{2178}{39601}\\\\(e_{6})^2=0.05499\\\\e_{6}=0.2345\\\\z_{6}=\frac{P'_{6}-P_{6}}{e_{6}}\\\\z_{6}=\frac{\frac{33}{199}-\frac{33}{199}}{0.2345}\\\\z_{6}=\frac{0}{46.6655}\\\\z_{6}=0

Result is not significant.

⇒If you will calculate the mean of all six z values, you will obtain that, z value is less than 2.So, we can say that ,outcomes are not equally likely at a 0.05 significance level.

⇒⇒Yes , as Probability of most of the numbers that is, 1,2,3,4,5,6 are different, for a loaded die , it should be equal to approximately equal to 33 for each of the numbers from 1 to 6.So, we can say with certainty that loaded die behaves differently than a fair​ die.

   

8 0
3 years ago
What is the chart called to convert decimals to a fraction
umka2103 [35]
Place value chart can be extended to include numbers less than one,which are sometimes called decimal fractions
4 0
3 years ago
Help!!
UNO [17]

Answer: financing a car

Step-by-step explanation:

i just took the apex test

5 0
3 years ago
Read 2 more answers
Other questions:
  • Is this linear or not linear? Please explain how you got your answer.
    8·2 answers
  • Shania is making a scale diagram of the badminton court at the community center. She uses a scale of 1 centimeter to 0.5 meter t
    5·1 answer
  • the senior class has 350 students. How much is the total bill if the student government wants to buy each senior a T-shirt costi
    12·1 answer
  • Find 0.01 more than 9.154
    13·1 answer
  • Find the sum of 1/3+2/3
    9·2 answers
  • Find the unknown values in the table below
    11·2 answers
  • I will upvote and give brainliest answer
    5·1 answer
  • Solve for the sides. WIll give brainliest if done in the next 10 minutes and is right
    12·1 answer
  • B) -2x + (5 - y)2 for x = -3 and y = 2
    8·1 answer
  • What is the greatest possible error if irina measured the length of her window as 3.35 feet? 0.5 0.05 0.005 0.0005
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!