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poizon [28]
3 years ago
8

Am i right?

Mathematics
1 answer:
lesantik [10]3 years ago
5 0
Yep you are 100% correct nice job yamiyugi1234
You might be interested in
. Which numbers are all divisible by 5? 1. 124, 333, 315, 266, 391 2. 135, 205, 330, 275, 365 3. 135, 211, 330, 274, 252 4. 232,
Ad libitum [116K]
Any numbers are ending with 5 and 0 can divide by 5
so answer:
<span>2. 135, 205, 330, 275, 365</span>
6 0
3 years ago
Read 2 more answers
HELP PLEASE ASAP
olga2289 [7]

Answer: a) degree and sign

b) end behavior: left side → +∞, right side → -∞

c) x-intercepts: x = -1.3, 0.3, 1.0

<u>Step-by-step explanation:</u>

end behavior can be determined by two things:

1) the degree of the polynomial:

  • if the degree is an even number, then the end behavior will be the same for both the left and right sides.
  • if the degree is an odd number, then the end behavior will be different for both the left and right sides.

2) the sign of the leading coefficient:

  • If the leading coefficient is positive, then the end behavior of the right side goes to positive infinity
  • If the leading coefficient is negative, then the end behavior of the right side goes to negative infinity

W(x) = -5x³ + 7x - 2

Degree: 3 (odd)

Leading Coefficient: negative

So, end behavior is: right side goes to negative infinity, right side goes to positive infinity.


See attachment for x-intercepts. <em>I set the x-axis to represent tenths </em>

5 0
3 years ago
If you flip a fair coin 6 times, what is the probability that you will get exactly 4 tails
mars1129 [50]

Answer: So the number of outcomes with exactly 4 tails is 720/2/24 = 15. Finally we can now calculate the probability of getting exactly 4 tails in 6 coin tosses as 15/64 = 0.234 to 3 decimal places.

4 0
3 years ago
Read 2 more answers
One sample has a mean of and a second sample has a mean of . The two samples are combined into a single set of scores. What is t
Murrr4er [49]

Answer:

a) For this case we can use the definition of weighted average given by:

M = \frac{ \bar X_1 n_1 + \bar X_2 n_2}{n_1 +n_2}

And if we replace the values given we have:

M = \frac{8*4 + 16*4}{4+4}= 12

b) M = \frac{8*3 + 16*5}{3+5}= 13

c) M = \frac{8*5 + 16*3}{5+3}= 11

Step-by-step explanation:

Assuming the following question: "One sample has a mean of M=8 and a second sample has a mean of M=16 . The two samples are combined into a single set of scores.

a) What is the mean for the combined set if both of the original samples have n=4 scores "

For this case we can use the definition of weighted average given by:

M = \frac{ \bar X_1 n_1 + \bar X_2 n_2}{n_1 +n_2}

And if we replace the values given we have:

M = \frac{8*4 + 16*4}{4+4}= 12

b) what is the mean for the combined set if the first sample has n=3 and the second sample has n=5

Using the definition we have:

M = \frac{8*3 + 16*5}{3+5}= 13

c) what is the mean for the combined set if the first sample has n=5 and the second sample has n=3

Using the definition we have:

M = \frac{8*5 + 16*3}{5+3}= 11

7 0
3 years ago
Solve and graph the compound inequality:<br> 5C + 3 &lt; 28 and -4C – 2 14
LekaFEV [45]

5 C + 3 < 28 a - 4 C - 214

a>9c +31

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

28. 4

C >28a-21

------

9

4 0
3 years ago
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