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balu736 [363]
3 years ago
12

Anne runs 5 6 of a mile every day. How many days would she have to run to reach a total of 35 6 miles?

Mathematics
1 answer:
matrenka [14]3 years ago
4 0
35/6) / (5/6
35/6*6/5
210/30=7
It would take her 7 days to run 35/6 miles.

Hope this helps :)
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What GCF can be factored out of the trinomial?
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Answer:

hahahaha bcjapfpcwkcncidj

Step-by-step explanation:

kdbcowkshxiwjs

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2 years ago
How many points need to be removed from this graph so that it will be a function?
DiKsa [7]
3 points. Do the vertical line test.
7 0
1 year ago
in a standard casino dice game the roller wins on the first roll if he rolls a sum of 7 or 11. what is the probability of winnin
Alborosie

<u>Answer-</u>

<em>The probability of winning on the first roll is </em><em>0.22</em>

<u>Solution-</u>

As in the game of casino, two dice are rolled simultaneously.

So the sample space would be,

|S|=6^2=36

Let E be the event such that the sum of two numbers are 7, so

E = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}

|E|=56

\therefore P(E)=\dfrac{|E|}{|S|}=\dfrac{6}{36}

Let F be the event such that the sum of two numbers are 11, so

F = {(6,5), (5,6)}

|F|=2

\therefore P(F)=\dfrac{|F|}{|S|}=\dfrac{2}{36}

Now,

P(\text{sum is 7 or 11)}=P(E\ \cup\ F)=P(E)+P(F)=\dfrac{6}{36}+\dfrac{2}{36}=\dfrac{8}{36}=\dfrac{2}{9}=0.22

8 0
3 years ago
The mean percentange of a population of people eating out at least once a week is 57℅
Sidana [21]

Answer:

<u>The correct answer is B. between 56.45% and 57.55% </u>

Complete statement and question:

The mean percentage of a population of people eating out at least once a week is 57% with a standard deviation of 3.50%. Assume that a sample size of 40 people was surveyed from the population a significant number of times. In which interval will 68% of the sample means occur?

between 55.89% and 58.11%

between 56.45% and 57.55%

between 56.54% and 57.46%

between 56.07% and 57.93%

Source: brainly.com/question/1068489

Step-by-step explanation:

1. Let's review the information given to us to answer the question correctly:

Mean percentage of a population of people eating out at least once a week  = 57%

Standard deviation = 3.5%

Sample size = 40

Confidence level = 68%

2. In which interval will 68% of the sample means occur?

For answering this question, we should find out the standard deviation of the sample, using this formula:

Standard deviation of the sample = Standard deviation of the population/√Sample size

Standard deviation of the sample = 3.5/√40

Standard deviation of the sample = 3.5/6.32

Standard deviation of the sample = 0.55

Let's recall that a confidence level of 68% means that 68% of the sample data would have a value between the mean - 1 time the standard deviation of the sample and the mean  + 1 time the standard deviation of the sample. Thus:

57 - 1 * 0.55 = 57 - 0.55 = 56.45

57 + 1  * 0.55 = 57 + 0.55 = 57.55

<u>The correct answer is B. between 56.45% and 57.55% </u>

7 0
3 years ago
The volume of a gas "V" varies inversely with the pressure "P" put on it. If the volume is 360cm³ under a pressure of 20 kgcm2
Sav [38]

Answer:

15 Kg cm²

Step-by-step explanation:

Given that V varies inversely with P then the equation relating them is

V = \frac{k}{P} ← k is the constant of variation

To find k use the condition

V = 360, P = 20, then

360 = \frac{k}{20} ( multiply both sides by 20 )

k = 7200

V = \frac{7200}{P} ← equation of variation

When V = 480, then

480 = \frac{7200}{P} ( multiply both sides by P )

480P = 7200 ( divide both sides by 480 )

P = 15

3 0
3 years ago
Read 2 more answers
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