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ra1l [238]
3 years ago
7

Can someone solve an Algebra 2 question for me and show work please.

Mathematics
1 answer:
lapo4ka [179]3 years ago
6 0
X to the second =4×3
x=√(12)
x=3.4641016151
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Systems of equations, help please?
lord [1]
4x + 3y = 54 (1)
3x + 9y = 108 (2)

Multiply (1) by (-3)
-12x - 9y = -162 
3x + 9y = 108
---------------------add
-9x  =  -54
   x = 6

plug x = 6 into (1) to find y
4(6) + 3y = 54
24 + 3y = 54
3y = 30
  y = 10

Answer
(6 , 10)

Hope it helps.
3 0
3 years ago
Find the area of a segment formed by a chord 8" long in a circle with radius of 8"
AlekseyPX
r=8\\\\P=\pi r^2\\\\P=\pi\cdot8^2=64\pi
5 0
3 years ago
The slope intercept form of the equation of a line that passes through (-2,13) is y =5x-3what is the point slope form of the equ
Zigmanuir [339]

Answer:

y - 13 = 5(x + 2)

Step-by-step explanation:

The equation of a line in point- slope form is

y - b = m(x - a)

where m is the slope and (a, b) a point on the line

Given y = 5x - 3 in  slope- intercept form

with slope m = 5, then

m = 5 and (a, b) = (- 2, 13)

y - 13 = 5(x - (- 2)), that is

y - 13 = 5(x + 2) ← in point- slope form

6 0
3 years ago
The set of ordered pairs in the mapping below can be described as which of the following?
PIT_PIT [208]
Because the x numbers are used more than once it cannot be a function. As for the the relation if its going through the origin then it would be a relation. im going to go with D
4 0
2 years ago
The proportion of brown M&M's in a milk chocolate packet is approximately 14%. Suppose a package of M&M's typically cont
marshall27 [118]

Answer:

Kindly check explanation

Step-by-step explanation:

Given the scenario above :

A) State the random variable.

The random variable is the p opoertion of brown M&M's in a milk chocolate packet.

B.) Argue that this is a binomial experiment.

Each trial is independent for a total number of 52 trials with a set probability of success at 0.14

C) probability that 6 M&M's are brown:.

P(x) = nCx * p^x * (1-p)^(n-x)

p = 0.14 ; (1 - p) = 0.86 ; n = 52 ; x = 6

P(x = 6) = 52C6 × 0.14^6 × 0.86^46

= 20358520 × 0.00000752954 × 0.00097035078

= 0.1487

D) P(x =25)

P(x = 25) = 52C25 × 0.14^25 × 0.86^27

= 477551179875952 × 449.987958058*10^(-24) × 0.01703955245

= 0.00000000366

E) P(x = 52)

P(x = 52) = 52C52 × 0.14^52 × 0.86^0

= 1 × 3968.78758299*10^(-48) × 1

= 3968.78758299*10^(-48)

F) yes it would be unusual, because such probability is extremely low. However, if a huge or substantial number of trials such may occur

3 0
3 years ago
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