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dedylja [7]
3 years ago
13

Tony is saving money to buy a game. So far he has saved S18, which is two-thirds of the total cost of the game. How much does th

e game cost
Mathematics
1 answer:
Anna35 [415]3 years ago
7 0
The game cost $27! I did 18 times 3 then dived that by 2! 18*3= 54. 54 divided by 2=27
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What is the formula of the sequence 3, 6, 10, 15, 21?​
AleksAgata [21]

Answer:

Tn=12n(n+1)

Step-by-step explanation:

Explanation:

These are the triangular numbers - each term in the sequence being the sum of the first n positive integers:

T1=1=1

T2=3=1+2

T3=6=1+2+3

etc.

Notice that:

2Tn=(0)+1+(0)+2+...+(n−1)+(0)+n

2Tn+(0)+n+(n−1)+...+(0)+2+(0)+1

2Tn=(n+1)+(n+1)+...+(n+1)+(n+1)

2Tn=n(n+1)

So:

Tn=12n(n+1)

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ehidna [41]

Answer:

P(X>1000)

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(X>1000)=P(\frac{X-\mu}{\sigma}>\frac{1000-\mu}{\sigma})=P(Z>\frac{1000-950}{50})=P(z>1)

And we can find this probability using the complement rule and the normal standard table and we got:

P(z>1)=1-P(z

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution problem

Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

X \sim N(950,50)  

Where \mu=950 and \sigma=50

We are interested on this probability

P(X>1000)

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(X>1000)=P(\frac{X-\mu}{\sigma}>\frac{1000-\mu}{\sigma})=P(Z>\frac{1000-950}{50})=P(z>1)

And we can find this probability using the complement rule and the normal standard table and we got:

P(z>1)=1-P(z

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I don’t know how to do this
lina2011 [118]
What is the question you need help on?
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The answer is Yeah I think
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