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trasher [3.6K]
3 years ago
12

Jake scored 67 on the calculus midterm. If the final exam counts twice as much as the midterm​ exam, then for what range of scor

es on the final would Jake get an average between 81 and 92​? Both tests have a maximum of 100 points.
Mathematics
1 answer:
pychu [463]3 years ago
7 0
Must be greater than or equal to 86
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For f(x)=x^2,sketch the graph of g(x)=f(-5x+10)
DiKsa [7]

The graph of g(x) = f(-5x+10) is given in the figure.

<h3>What is a graph?</h3>

A diagram showing the relation between two variable quantities,each measured along one of a pair of axes at right angles.

It is given that f(x) = x^2

and g(x ) = f(-5x+10)

Now putting the value of f(x) in g(x) we get,

g(x)  = f(-5x+10)  = (-5x+10)^2

So, g(x) = (-5x+10)^2

now, making the table for g(x),

<u><em>x </em></u><u>g(x)</u>

0 100

1 81

2 0

3 25

4 100

5 225

Hence,the graph of g(x) = f(-5x+10) is given in the figure.

More about graph :

brainly.com/question/11616742

#SPJ1

3 0
1 year ago
What is the slope intercept of the equation of the line through the given point (0,3) and (-4,-1)
cestrela7 [59]

Answer:

y=x+3

Step-by-step explanation:

1.) Find slope

slope = (-1-3)/(-4-0)

= -4/-4

= 1

2.) Use point slope form

<u>y-y1= m (x-x1) </u>

both x1 and y1 are the points

m is the slope

3.) Plug in the numbers

for the points I chose (0,3) to plug in

y-3= 1 (x1-0)

y-3= x

y= x+3

8 0
3 years ago
15 points!!!
ankoles [38]

Answer:

Step-by-step explanation:

4/15×6/14= 0.1142

4 0
3 years ago
HELP PLEASE!!! I need help with 94 if you could show the steps that would be very helpful!
aksik [14]
A combination is an unordered arrangement of r distinct objects in a set of n objects. To find the number of permutations, we use the following equation:

n!/((n-r)!r!)

In this case, there could be 0, 1, 2, 3, 4, or all 5 cards discarded. There is only one possible combination each for 0 or 5 cards being discarded (either none of them or all of them). We will be the above equation to find the number of combination s for 1, 2, 3, and 4 discarded cards.

5!/((5-1)!1!) = 5!/(4!*1!) = (5*4*3*2*1)/(4*3*2*1*1) = 5

5!/((5-2)!2!) = 5!/(3!2!) = (5*4*3*2*1)/(3*2*1*2*1) = 10

5!/((5-3)!3!) = 5!/(2!3!) = (5*4*3*2*1)/(2*1*3*2*1) = 10

5!/((5-4)!4!) = 5!/(1!4!) = (5*4*3*2*1)/(1*4*3*2*1) = 5

Notice that discarding 1 or discarding 4 have the same number of combinations, as do discarding 2 or 3. This is being they are inverses of each other. That is, if we discard 2 cards there will be 3 left, or if we discard 3 there will be 2 left.

Now we add together the combinations

1 + 5 + 10 + 10 + 5 + 1 = 32 choices combinations to discard.

The answer is 32.

-------------------------------

Note: There is also an equation for permutations which is:

n!/(n-r)!

Notice it is very similar to combinations. The only difference is that a permutation is an ORDERED arrangement while a combination is UNORDERED.

We used combinations rather than permutations because the order of the cards does not matter in this case. For example, we could discard the ace of spades followed by the jack of diamonds, or we could discard the jack or diamonds followed by the ace of spades. These two instances are the same combination of cards but a different permutation. We do not care about the order.

I hope this helps! If you have any questions, let me know :)








7 0
3 years ago
Please help with geometry homework !!!
Umnica [9.8K]

You can solve this with next proportion =>

3 : 5 = 4 : x => 3x= 5*4 => 3x = 20 => x=20/3 = 6 (2/3)

The correct answer is  x= 6 (2/3)

Good luck!!!

5 0
3 years ago
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