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4 years ago

Help finding mean 14 19 13 22 7

Mathematics

2 answers:

bazaltina [42]4 years ago

7 0

The answer is 15. you add up all the numbers, and divide it by 5 because thats how many numbers there are. Pretty basic ;)

Elis [28]4 years ago

5 0

You just have to add all the number together and divide it by the amount of the number so it would equal 15

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I need Help ASAP I will give you brainlest and plus 5 stars ✨✨and plus thanks and 50 points EVeryONe need help this is due today

murzikaleks [220]

Answer:

see step-by-step below

Step-by-step explanation:

1) y - 2 > 3

Add 2 to both sides: y > 5

2) y ≤ 2

Draw a closed circle at y = 2 then shade the numbers to the left

3) y ≥ 250

where y = savings

4) The image isn't very clear, so I can't tell if it's a closed or open circle on -4.

If it is closed, it will be y ≤ -4

If it is open, it will be y < -4

5) -4 < f

Draw an open circle at f = -4 then shade the numbers to the right

6) x < -3.5

Draw an open circle at x = -3.5 then shade the numbers to the left

7) closed circle ... shade to the left of 8 ... make y equal to or less than 8.

8) F ≤ 6

C - closed circle at 6 with shaded numbers to the left

9) y > 2.75

Betty saved more than $2.75 every month

10) 1 + x < 5

Subtract 1 from both sides: x < 4

11) x ≥ -2.5

12) 1 ≤ m means that m is equal to or greater than 1

solutions: 4, 1, 2.3

3 0

2 years ago

Please help i will give brainliest.

fredd [130]

Answer:

It’s a pair of : corresponding angles

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3 years ago

Read 2 more answers

Ella began doing her homework at a quarter five in the afternoon she finished her homework at 5:35 how long did it take Ella to

Alisiya [41]

Answer: 50 minutes

Step-by-step explanation: a quarter to 5 is always 15 minutes from 5, 4:45. So 5:35-4:45 is 50.

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3 years ago

Prove that if $w,z$ are complex numbers such that $|w|=|z|=1$ and $wz\ne -1$, then $\frac{w+z}{1+wz}$ is a real number.

mr_godi [17]

Answer:

See proof below

Step-by-step explanation:

Let $r=\frac{w+z}{1+wz}$ . If w=-z, then r=0 and r is real. Suppose that w≠-z, that is, r≠0.

Remember this useful identity: if x is a complex number then $x\bar{x}=|x|^2$ where $\bar{x}$ is the conjugate of x.

Now, using the properties of the conjugate (the conjugate of the sum(product) of two numbers is the sum(product) of the conjugates):

$\frac{r}{\bar{r}}=\frac{w+z}{1+wz} \left(\frac{1+\bar{w}\bar{z}}{\bar{w}+\bar{z}}{\right)$

= $\frac{(w+z)(1+\bar{w}\bar{z})}{(1+wz)(\bar{w}+\bar{z})}=\frac{w+z+w\bar{w}\bar{z}+z\bar{z}\bar{w}}{\bar{w}+\bar{z}+\bar{w}wz+\bar{z}zw}=\frac{w+z+w+|w|^2\bar{z}+|z|^2\bar{w}}{\bar{w}+\bar{z}+|w|^2z+|z|^2w}=\frac{w+z+\bar{z}+\bar{w}}{\bar{w}+\bar{z}+z+w}=1$