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Bumek [7]
3 years ago
6

Simplify the expression: -2.3f+0.8f - 12-5

Mathematics
1 answer:
MaRussiya [10]3 years ago
8 0

Answer:

-1.5f - 17

Step-by-step explanation:

-2.3f + 0.8f - 12 - 5

= -1.5f - 17

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Ella has a points card for a movie theater.
MAXImum [283]

Answer:

175<=12.5x+40

Step-by-step explanation:

She starts with 40 points already so you add in those 40 then since she gets 12.5 points per visit that is your x, so you set that as greater than or equal to 175

7 0
3 years ago
Please please help!!!! According to the probability distribution below, what is the expected value of X? X 0 1 2 3 4 5 P(X) 0.3
Novay_Z [31]

Answer:

D.) The expected value is 1.78

Step-by-step explanation:

The expected value of a probability distribution is evaluated using the formula.

Expected Value, E(X)=\sum^{n}_{i=1}x_iP(x_i)

Therefore, from the given probability distribution, we have:

E(X)=(0*0.3)+(1*0.2)+(2*0.16)+(3*0.2)+(4*0.04)+(5*0.1)

E(X)=1.78

The Expected value of X is 1.78.

The correct option is D.

4 0
3 years ago
What is 4.2345 rounded to the nearest tenth
MAVERICK [17]

Answer to question:

4.2

7 0
3 years ago
Read 2 more answers
Brainliest,plz help
labwork [276]

4 sqrt (32) + 6 sqrt (50)

4 sqrt (16*2) + 6 sqrt (2*25)

4 sqrt (16)*sqrt(2) + 6 sqrt (2)*sqrt(25)

4 *4*sqrt(2)+6sqrt(2)*5

16sqrt(2)+30sqrt(2)

46sqrt(2)


4 0
3 years ago
for any positive integer n, the sum of the first n positive integers equals n(n 1)/2. what is the sum of all the even integers b
marusya05 [52]

The sum of all the even integers between 99 and 301 is 20200

To find the sum of even integers between 99 and 301, we will use the arithmetic progressions(AP). The even numbers can be considered as an AP with common difference 2.

In this case, the first even integer will be 100 and the last even integer will be 300.

nth term of the AP = first term + (n-1) x common difference

               ⇒    300 = 100 + (n-1) x 2

Therefore, n = (200 + 2 )/2 = 101

That is, there are 101 even integers between 99 and 301.

Sum of the 'n' terms in an AP = n/2 ( first term + last term)

                                                = 101/2 (300+100)

                                                = 20200

Thus sum of all the even integers between 99 and 301 = 20200

Learn more about arithmetic progressions at brainly.com/question/24592110

#SPJ4        

7 0
1 year ago
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