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Anna11 [10]
3 years ago
14

Tyler needs at least $205 for a new video game system. He has already saves $30. He earns $7 an hour at his job. Write and solve

an inequality to find how many hours he will need to work to buy the system. Interpret the solution.
Mathematics
2 answers:
azamat3 years ago
5 0
Amount needed=earned+already have
205=earned+30
earned=7 times number of hours=7h

205≤7h+30
we used the less than or equal to because he needs at least 205, means 205 or greater must be earned


205≤7h+30
minus 30 both sides
175≤7h
divide both sides by 7
25≤h
needs to work at least 25 hours
vesna_86 [32]3 years ago
3 0
So, first, gather your important information:

"at least $205"
"already has $30"
"$7 per hour"

now rewrite this into numbers and symbols:

"at least $205" ... >/= 205 (his money MUST be over 205)
"already has $30" ... +30 (he's already saved 30, so add it)
"$7 per hour" ... 7h (he makes 7 an hour)

put all of those together.
7h + 30 >/= 205

this is your inequality. seven dollars an hour plus the thirty dollars he already saved must equal at least 205.

now, solve it for h:
7h + 30 >/= 205 ... subtract 30
7h >/= 175 ... divide by 7
h >/= 25

because your h variable stands for the hours worked, he must work at least 25 hours to have enough money to buy his new video game system.
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Step-by-step explanation:

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andreev551 [17]

g(x) is a shift of 8 units to the left and 4 units of f(x), then the correct statement is B.

<h3>Which statement compares the graph of the two functions?</h3>

First, a vertical shift of N units is written as:

g(x) = f(x) + N

  • if N > 0 the shift is upwards.
  • If N < 0 the shift is downwards.

A horizontal shift of N units is written as:

g(x) = f(x + N).

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In this case, we have:

g(x) = f(x + 8) + 4

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8 0
1 year ago
Grace wants to find the probability of a family of 3 children having 2 boys and 1 girl. Use a tree diagram to list the possibili
pychu [463]

Answer:

Step-by-step explanation:

My approach was to draw out the probabilities, since we have 3 children, and we are looking for 2 boys and 1 girl, the probabilities can be Boy-Boy-Girl, Boy-Girl-Boy, and Girl-Boy-Boy. So a 2/3 chance if you think about it, your answer 2/3 can't be correct. If we assume that boys and girls are born with equal probability, then the probability to have two girls (and one boy) should be the same as the probability to have two boys and one girl. So you would have two cases with probability 2/3, giving an impossible 4/3 probability for both cases. Also, your list "Boy-Boy-Girl, Boy-Girl-Boy, and Girl-Boy-Boy" seems strange. All of those are 2 boys and 1 girl, so based on that list, you should get a 100 percent chance. But what about Boy-Girl-Girl, or Girl-Girl-Girl? You get 2/3 if you assume that adjacencies in the (ordered) list are important, i.e., "2 boys and a girl" means that the girl was not born between the boys.

5 0
1 year ago
A worker was paid a salary of $10,500 in 1985. Each year, a salary increase of 6% of the previous year's salary was awarded. How
Mazyrski [523]
Note that 6% converted to a decimal number is 6/100=0.06. Also note that 6% of a certain quantity x is 0.06x.

Here is how much the worker earned each year:


In the year 1985 the worker earned <span>$10,500. 

</span>In the year 1986 the worker earned $10,500 + 0.06($10,500). Factorizing $10,500, we can write this sum as:

                                            $10,500(1+0.06).



In the year 1987 the worker earned

$10,500(1+0.06) + 0.06[$10,500(1+0.06)].

Now we can factorize $10,500(1+0.06) and write the earnings as:

$10,500(1+0.06) [1+0.06]=$10,500(1.06)^2.


Similarly we can check that in the year 1987 the worker earned $10,500(1.06)^3, which makes the pattern clear. 


We can count that from the year 1985 to 1987 we had 2+1 salaries, so from 1985 to 2010 there are 2010-1985+1=26 salaries. This means that the total paid salaries are:

10,500+10,500(1.06)^1+10,500(1.06)^2+10,500(1.06)^3...10,500(1.06)^{26}.

Factorizing, we have

=10,500[1+1.06+(1.06)^2+(1.06)^3+...+(1.06)^{26}]=10,500\cdot[1+1.06+(1.06)^2+(1.06)^3+...+(1.06)^{26}]

We recognize the sum as the geometric sum with first term 1 and common ratio 1.06, applying the formula

\sum_{i=1}^{n} a_i= a(\frac{1-r^n}{1-r}) (where a is the first term and r is the common ratio) we have:

\sum_{i=1}^{26} a_i= 1(\frac{1-(1.06)^{26}}{1-1.06})= \frac{1-4.55}{-0.06}= 59.17.



Finally, multiplying 10,500 by 59.17 we have 621.285 ($).


The answer we found is very close to D. The difference can be explained by the accuracy of the values used in calculation, most important, in calculating (1.06)^{26}.


Answer: D



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3 years ago
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