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Jlenok [28]
3 years ago
12

Finding changed average

Mathematics
2 answers:
allsm [11]3 years ago
8 0

Answer:

15.

Step-by-step explanation:

The new average will be 11 + 4 = 15.

LekaFEV [45]3 years ago
5 0

Answer: The overall average will increase by 4 from 11 to 15

Step-by-step explanation: If each of the six integers is increased by four, that will add 6×4 to the total.

The original average was based on six numbers totaling 66, then divided by 6 to get the average, 11.

So add 66 + 24 = 90

Divide 90/6 = 15. The new average.

That is 4 more than the original 11.

You might be interested in
The sum of the angles of a triangle is 180°. If one angle of a triangle measures x and the second angle measures (5x +19)º
lapo4ka [179]

Answer:

third angle = (-6x + 161)°

Step-by-step explanation:

180° = (x)° + (5x + 19)° + third angle →

180° = (6x + 19)° + third angle →

180° – (6x + 19)° = (6x + 19)° + third angle – (6x + 19)° →

180° – (6x + 19)° = third angle →

180° + -(6x + 19)° = third angle →

180° + (-6x)° + (-19)° = third angle

180° – 6x° – 19° = third angle

(180° – 19°) – 6x° = third angle

161° – 6x° = third angle

third angle = 161 – 6x°

third angle = -6x° + 161°

third angle = (-6x + 161)°

8 0
3 years ago
Which of these points, if any, is the midpoint of FJ?
ivanzaharov [21]

The mid-point of FJ is the number right in the middle of FJ. Find the mean of F & J. Add regardless of sign

4 + 6 = 10

10/2 = 5

Add 5 to -4

5 + -4 = 1

H is your midpoint

True. As long as AB and BC are on the same line, AB + BC = AC

hope this helps

4 0
3 years ago
Solve by substitution y=x+4, y=2x+5
Amiraneli [1.4K]
X+4=2x+5
-1=x
hope this helps
4 0
2 years ago
I need help with part b. I feel like there’s a catch, I want to do the first derivative test, however, I feel like there is a be
Sladkaya [172]

Answer:

The fifth degree Taylor polynomial of g(x) is increasing around x=-1

Step-by-step explanation:

Yes, you can do the derivative of the fifth degree Taylor polynomial, but notice that its derivative evaluated at x =-1 will give zero for all its terms except for the one of first order, so the calculation becomes simple:

P_5(x)=g(-1)+g'(-1)\,(x+1)+g"(-1)\, \frac{(x+1)^2}{2!} +g^{(3)}(-1)\, \frac{(x+1)^3}{3!} + g^{(4)}(-1)\, \frac{(x+1)^4}{4!} +g^{(5)}(-1)\, \frac{(x+1)^5}{5!}

and when you do its derivative:

1) the constant term renders zero,

2) the following term (term of order 1, the linear term) renders: g'(-1)\,(1) since the derivative of (x+1) is one,

3) all other terms will keep at least one factor (x+1) in their derivative, and this evaluated at x = -1 will render zero

Therefore, the only term that would give you something different from zero once evaluated at x = -1 is the derivative of that linear term. and that only non-zero term is: g'(-1)= 7 as per the information given. Therefore, the function has derivative larger than zero, then it is increasing in the vicinity of x = -1

6 0
3 years ago
NEED HELP, LIMITED TIME, HURRY
11111nata11111 [884]

Answer:

9

Step-by-step explanation:

here you go!

6 0
3 years ago
Read 2 more answers
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