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Maurinko [17]
3 years ago
12

I am graphing something with 29 points. I know that you have to label the points with a capital letter. However, there are only

26 letters in the English alphabet, so what do I name the remaining 3 points?​

Mathematics
1 answer:
allochka39001 [22]3 years ago
6 0

Answer:

a₁, a₂, a₃, . . .

Step-by-step explanation:

When we name something in Mathematics it is always advisable to number them as $ a_1, a_2, a_3, ...$.

Because the alphabets are only 26 in number we might run out of notations.

The simple logic behind the notion of $ a_1, a_2, a_3, ... $ is that the numbers have no end and we can number as many variables we want using this logic.

In this problem, you can continue your notation from $ a_{27}, a_{28}, a_{29} $ so that you don't have to make a change in your figure.

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3 years ago
Help! i need to answer this before 11:59pm EST, Today !!, i would appreciate your help.
Arada [10]

Answer:

0.18 sec or 3.4 sec.

For second part, it will take 3.7 sec.

First part:

Set h(t) = 17and solve for t.

-16t²+ 58t + 7= 17

-16t² + 58t - 10 = 0

Solve this quadratic equation for t. You should get 2 positive solutions. The lower value is the time to reach 17 on the way up, and the higher value is the time to reach 17 again, on the way down.

Second part:

Set h(t) = 0 and solve the resulting quadratic equation for t. You should get a negative solution (which you can discard), and a positive solution. The latter is your answer.

3 0
4 years ago
Y = 119.67 (0.61) ÷ 5
Travka [436]
Y=  119.67 (0.61) <span>÷ 5
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4 0
3 years ago
Tara has two standard, six-sided dice. If she rolls them 36 times, how many times should she expect a sum less than 5 to show up
andreyandreev [35.5K]
There are six sides on each die. For each possible score on Die 1, there are six possible scores on Die 2. So the number of possible combinations is 6*6 = 36. 
<span>It follows that if the dice are thrown 36 times, you would expect each combination to come up once. </span>

<span>We therefore simply need to know how many combinations add up to less than 5. (I've interpreted this as not including a total of 5 itself). </span>
<span>These combinations are: 1 and 1, 2 and 1, 1 and 2, 2 and 2, 3 and 1, and 1 and 3 ---> six combinations out of 36. </span>

<span>So you'd expect a sum less than 5 six times. </span>
8 0
4 years ago
A number is picked randomly in the range [0,7]. What is the probability that the number picked is between 3 and 5? What is the p
emmasim [6.3K]

Answer:

(1) 0.125

(2) 0.125

Step-by-step explanation:

The total number of possible outcomes is:

N = 8

(1)

Compute the probability that the number picked is between 3 and 5 as follows:

Number of Favorable outcomes = 1

The probability is:

P (Number picked is between 3 and 5) = 1/8 = 0.125

Thus, the probability that the number picked is between 3 and 5 is 0.125.

(2)

The number usually picked appears to be in in the range [3,5], i.e. the numbers could be, {3, 4 or 5}.

Number of Favorable outcomes = n (Number < 4 and within [3, 5]) = 1

P (less than 4 ∩ within [3, 5]) = 1/8 = 0.125

Thus, the probability that the number picked is less than 4 knowing that the number usually picked appears to be in in the range [3,5] is 0.125.

4 0
3 years ago
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