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Nana76 [90]
3 years ago
14

Jenny has five exam scores of 79,66, 71, 89, and 84 in her biology class. What score does she need on the final exam to have a m

ean grade of 80?
Round your answer to two decimal places, if necessary. (All exams have a maximum of 100 points.)
Mathematics
2 answers:
olganol [36]3 years ago
5 0

Answer:

she needs to get a 99

Step-by-step explanation:

andrezito [222]3 years ago
5 0

Answer: 76.34

Step-by-step explanation:

79.66 + 71 + 89 + 84 + 76.34 = 400/5 = 80

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Find the number of possible ways one can choose marbles from a bag containing three marbles: a) if the order is important b) if
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Answer:

There are 3 marbles, that we assume are different.

Let's define them as marble 1, marble 2, and marble 3.

a) If the order is important, and we select the 3 marbles, then:

for the first marble, we have 3 options.

For the second marble, we have 2 options (because one is already taken)

for the third marble is only one option.

The total number of combinations is equal to the product between the numbers of options, so in this case, we have:

C = 3*2*1 = 6 combinations.

Now, if we only select 2 marbles from the bag, we have:

for the first marble, we have 3 options.

For the second marble, we have 2 options

Here the number of different combinations is:

C = 3*2 = 6 combinations.

And if we select only one marble from the bag, we have:

for the first marble, we have 3 options.

Then here are only 3 combinations.

the total number of combinations is:

C' = 6 + 6 + 3 = 15 different combinations.

b) If the order is not important, then there is only one combination when we draw the 3 marbles.

When we draw one marble there are 3 combinations (one for each marble)

When we draw two marbles, again there are 3 combinations (one for each marble that we do not draw)

then the total number of combinations in this case is:

C = 1 + 3 + 3 = 7

3 0
3 years ago
What's the answer? I need help!
Juli2301 [7.4K]

Answer:

<h2><u>I can do algebra forever</u></h2>

First we can use distributive property. 2a-10 = 6. So we add 10 on both sides and get 2a=16. Then we divide both sides by 2 so we get a = 8. Therefore proved algebraically that a = 8. I can always do algebra. Feel free to keep asking questions like these. Well feel free to ask any question because this is brainly.

<h2><u>Answer is a = 8, 8 is answer</u></h2><h2><u></u></h2>

<u>brainliest</u>

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3 years ago
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16. How many different words can be formed with the letters of the word 'RAJARAM'? In how many of
Mandarinka [93]

Answer:

The word "ARRANGE" can be arranged in

2!×2!

7!

=

4

5040

=1260 ways.

For the two R's do occur together, let us make a group of R's taking from "ARRANGE" and permute them.

Then the number of ways =

2!

6!

=360.

The number ways to arrange "ARRANGE", where two "R's" will not occur together is =1260−360=900.

Also in the same way, the number of ways where two "A's" are together is 360.

The number of ways where two "A's" and two "R's" are together is 5!=120.

The number of ways where neither two "A's" nor two "R's" are together is =1260−(360+360)+120=660.

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