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Bess [88]
3 years ago
7

Solve for x: 2|x − 3| + 1 = 7

Mathematics
1 answer:
77julia77 [94]3 years ago
8 0

Answer:

X=6

Step-by-step explanation:

2(x-3) +1=7

2x -6+1=7

2x=7+6-1

2x=12

2x÷2=12÷2

X=6

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A lawn had an area of 100 square feet. If it was 10 feet width, how long was it?
AleksAgata [21]
10 ft because 10×10=100
3 0
3 years ago
Read 2 more answers
A poll agency reports that 75% of teenagers aged 12-17 own smartphones. A random sample of 234 teenagers is drawn. Round your an
GalinKa [24]

Answer:

If our random variable of interest for this case is X="the number of teenagers between 12-17 with smartphone" we can model the variable with this distribution:

X \sim Binom(n=234, p=0.75)

And the mean for this case would be:

E(X) =np = 234*0.75= 175.5

And the standard deviation would be given by:

\sigma =\sqrt{np(1-p)}= \sqrt{234*0.75*(1-0.75)}= 6.624

Step-by-step explanation:

If our random variable of interest for this case is X="the number of teenagers between 12-17 with smartphone" we can model the variable with this distribution:

X \sim Binom(n=234, p=0.75)

And the mean for this case would be:

E(X) =np = 234*0.75= 175.5

And the standard deviation would be given by:

\sigma =\sqrt{np(1-p)}= \sqrt{234*0.75*(1-0.75)}= 6.624

3 0
3 years ago
Please help answer these
BabaBlast [244]
You should use the ratio so 

2/ 4  = 6.5 / n  

n= 6.5 x 4  / 2 =  13 inches 


2- for two hours ------> four inches of snow 
for x hours -----> y inches of snow 
the equation is 
2/4 = x/ y 

x= 2y / 4 
 for x is the hours and y is the inches 

exercice 2
5 gr granula ----------------------> 1750 calories 
2 gr granula ----------------------> n ?? calories 

by cross multiplication 
n = 2 x1750 / 5 =  700 calories .

2-
she wants to eat 175 calories so 
u put granula under granula  and calories under calories to make this cross 

<span>5 gr granula ----------------------> 1750 calories 
</span>??? m granula ---------------------> 175 calories 
<span>by cross multiplication </span>
m = 175 x 5 / 1750 =  0.5 gr of granula 
8 0
4 years ago
How do you do these two questions?
nignag [31]

Step-by-step explanation:

(a) ∫₋ₒₒ°° f(x) dx

We can split this into three integrals:

= ∫₋ₒₒ⁻¹ f(x) dx + ∫₋₁¹ f(x) dx + ∫₁°° f(x) dx

Since the function is even (symmetrical about the y-axis), we can further simplify this as:

= ∫₋₁¹ f(x) dx + 2 ∫₁°° f(x) dx

The first integral is finite, so it converges.

For the second integral, we can use comparison test.

g(x) = e^(-½ x) is greater than f(x) = e^(-½ x²) for all x greater than 1.

We can show that g(x) converges:

∫₁°° e^(-½ x) dx = -2 e^(-½ x) |₁°° = -2 e^(-∞) − -2 e^(-½) = 0 + 2e^(-½).

Therefore, the smaller function f(x) also converges.

(b) The width of the intervals is:

Δx = (3 − -3) / 6 = 1

Evaluating the function at the beginning and end of each interval:

f(-3) = e^(-9/2)

f(-2) = e^(-2)

f(-1) = e^(-1/2)

f(0) = 1

f(1) = e^(-1/2)

f(2) = e^(-2)

f(3) = e^(-9/2)

Apply Simpson's rule:

S = Δx/3 [f(-3) + 4f(-2) + 2f(-1) + 4f(0) + 2f(1) + 4f(2) + f(3)]

S ≈ 2.5103

5 0
3 years ago
PLEASE HELP????????!!!​
kirza4 [7]

Answer:

7.7i think if that's even an answer hope this helps

5 0
3 years ago
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