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Margarita [4]
3 years ago
11

I need some help... Algebra Hw

Mathematics
1 answer:
Svet_ta [14]3 years ago
4 0

Answer:can't see no 1

Step-by-step explanation:

2=A

3=c

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Roman55 [17]

Answer:

Three (3) is the correct answer

4 0
3 years ago
Need help pls and thank you
Mkey [24]

Answer:

y=25x

Step-by-step explanation:

It is a linear equation with a slope of 25

5 0
2 years ago
Can u solve this for me: 72 + 4x = 2x + 82 and please show work
8_murik_8 [283]
The answer is X = 5 because first you have to take away 2x from that side so u minus 2x from 4x that gives u: 72 + 2x = 82 then you have to minus 72 from both sides and that gives you 2x = 10 once you have that you divide both sides by 2 because you want the X by itself and once you get that X = 5. Hope this helped!
8 0
3 years ago
Read 2 more answers
Find the area under the standard normal probability distribution between the following pairs of​ z-scores. a. z=0 and z=3.00 e.
prohojiy [21]

Answer:

a. P(0 < z < 3.00) =  0.4987

b. P(0 < z < 1.00) =  0.3414

c. P(0 < z < 2.00) = 0.4773

d. P(0 < z < 0.79) = 0.2852

e. P(-3.00 < z < 0) = 0.4987

f. P(-1.00 < z < 0) = 0.3414

g. P(-1.58 < z < 0) = 0.4429

h. P(-0.79 < z < 0) = 0.2852

Step-by-step explanation:

Find the area under the standard normal probability distribution between the following pairs of​ z-scores.

a. z=0 and z=3.00

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 3.00) = 0.9987

Thus;

P(0 < z < 3.00) = 0.9987 - 0.5

P(0 < z < 3.00) =  0.4987

b. b. z=0 and z=1.00

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 1.00) = 0.8414

Thus;

P(0 < z < 1.00) = 0.8414 - 0.5

P(0 < z < 1.00) =  0.3414

c. z=0 and z=2.00

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 2.00) = 0.9773

Thus;

P(0 < z < 2.00) = 0.9773 - 0.5

P(0 < z < 2.00) = 0.4773

d.  z=0 and z=0.79

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 0.79) = 0.7852

Thus;

P(0 < z < 0.79) = 0.7852- 0.5

P(0 < z < 0.79) = 0.2852

e. z=−3.00 and z=0

From the standard normal distribution tables,

P(Z< -3.00) = 0.0014  and P(Z< 0) = 0.5

Thus;

P(-3.00 < z < 0 ) = 0.5 - 0.0013

P(-3.00 < z < 0) = 0.4987

f. z=−1.00 and z=0

From the standard normal distribution tables,

P(Z< -1.00) = 0.1587  and P(Z< 0) = 0.5

Thus;

P(-1.00 < z < 0 ) = 0.5 -  0.1586

P(-1.00 < z < 0) = 0.3414

g. z=−1.58 and z=0

From the standard normal distribution tables,

P(Z< -1.58) = 0.0571  and P(Z< 0) = 0.5

Thus;

P(-1.58 < z < 0 ) = 0.5 -  0.0571

P(-1.58 < z < 0) = 0.4429

h. z=−0.79 and z=0

From the standard normal distribution tables,

P(Z< -0.79) = 0.2148  and P(Z< 0) = 0.5

Thus;

P(-0.79 < z < 0 ) = 0.5 -  0.2148

P(-0.79 < z < 0) = 0.2852

8 0
3 years ago
Find the value of<br> 10!<br> (10-3)<br> A. 70<br> B. 120<br> O C. 5040<br> O D. 720
GarryVolchara [31]

Answer:

The answer is option D.

Step-by-step explanation:

\frac{10!}{(10 - 3) ! }  =  \frac{10!}{7!}  =  \frac{3628800}{5040}  = 720

Hope this helps you

8 0
3 years ago
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