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lorasvet [3.4K]
1 year ago
15

Calculate (3.012 x 10−8) + (4 x 10−8). 3.016 x 10−16 7.012 x 10−16 7.012 x 10−8 7.012 x 20−8

Mathematics
1 answer:
Kobotan [32]1 year ago
8 0

Answer: -3350.104 because you need to multiply the brackets first then subtract.

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F(x)=3x2+2x−3 <br><br> g(x)=2x+4
Yakvenalex [24]

Answer:

A: 6

B: 16

C: 2

D: -12

Step-by-step explanation:

They are asking you to multiply the equations:

(3x^2 + 2x - 3)(2x + 4)

Distribute and you should get:

6x^3 + 16x^2 + 2x - 12

The Coefficients A,B,C, and D are 6, 16, 2, and -12.

3 0
3 years ago
What is the cost of gas at 4.6 $2.70 per gallon
Anarel [89]

Answer:

$12.42

Step-by-step explanation:

4.6 gallons at 2.70 per gallon.

4.6*2.70= $12.42

3 0
3 years ago
Find the value of x - Secant and Tangent Angles in Circles
VladimirAG [237]

Answer:

  C.  70°

Step-by-step explanation:

The inscribed angle marked 15° intercepts an arc that is double that measure, so the intercepted arc on the right is 2×15° = 30°.

The external angle marked 20° is half the difference of the intercepted arcs, so is ...

  20° = (1/2)(x - 30°)

  40° = x - 30° . . . . . . multiply by 2

  70° = x . . . . . . . . . . . add 30°

The value of x is 70°.

6 0
3 years ago
11. Let X denote the amount of time a book on two-hour reserve is actually checked out, and suppose the cdf is F(x) 5 5 0 x , 0
NISA [10]

Question not properly presented

Let X denote the amount of time a book on two-hour reserve is actually checked out, and suppose the cdf is F(x)

0 ------ x<0

x²/25 ---- 0 ≤ x ≤ 5

1 ----- 5 ≤ x

Use the cdf to obtain the following.

(a) Calculate P(X ≤ 4).

(b) Calculate P(3.5 ≤ X ≤ 4).

(c) Calculate P(X > 4.5)

(d) What is the median checkout duration, μ?

e. Obtain the density function f (x).

f. Calculate E(X).

Answer:

a. P(X ≤ 4) = 16/25

b. P(3.5 ≤ X ≤ 4) = 3.75/25

c. P(4.5 ≤ X ≤ 5) = 4.75/25

d. μ = 3.5

e. f(x) = 2x/25 for 0≤x≤2/5

f. E(x) = 16/9375

Step-by-step explanation:

a. Calculate P(X ≤ 4).

Given that the cdf, F(x) = x²/25 for 0 ≤ x ≤ 5

So, we have

P(X ≤ 4) = F(x) {0,4}

P(X ≤ 4) = x²/25 {0,4}

P(X ≤ 4) = 4²/25

P(X ≤ 4) = 16/25

b. Calculate P(3.5 ≤ X ≤ 4).

Given that the cdf, F(x) = x²/25 for 0 ≤ x ≤ 5

So, we have

P(3.5 ≤ X ≤ 4) = F(x) {3.5,4}

P(3.5 ≤ X ≤ 4) = x²/25 {3.5,4}

P(3.5 ≤ X ≤ 4) = 4²/25 - 3.5²/25

P(3.5 ≤ X ≤ 4) = 16/25 - 12.25/25

P(3.5 ≤ X ≤ 4) = 3.75/25

(c) Calculate P(X > 4.5).

Given that the cdf, F(x) = x²/25 for 0 ≤ x ≤ 5

So, we have

P(4.5 ≤ X ≤ 5) = F(x) {4.5,5}

P(4.5 ≤ X ≤ 5) = x²/25 {4.5,5}

P(4.5 ≤ X ≤ 5)) = 5²/25 - 4.5²/25

P(4.5 ≤ X ≤ 5) = 25/25 - 20.25/25

P(4.5 ≤ X ≤ 5) = 4.75/25

(d) What is the median checkout duration, μ?

Median is calculated as follows;

∫f(x) dx {-∝,μ} = ½

This implies

F(x) {-∝,μ} = ½

where F(x) = x²/25 for 0 ≤ x ≤ 5

F(x) {-∝,μ} = ½ becomes

x²/25 {0,μ} = ½

μ² = ½ * 25

μ² = 12.5

μ = √12.5

μ = 3.5

e. Calculating density function f (x).

If F(x) = ∫f(x) dx

Then f(x) = d/dx (F(x))

where F(x) = x²/25 for 0 ≤ x ≤ 5

f(x) = d/dx(x²/25)

f(x) = 2x/25

When

F(x) = 0, f(x) = 2(0)/25 = 0

When

F(x) = 5, f(x) = 2(5)/25 = 2/5

f(x) = 2x/25 for 0≤x≤2/5

f. Calculating E(X).

E(x) = ∫xf(x) dx, 0,2/5

E(x) = ∫x * 2x/25 dx, 0,2/5

E(x) = 2∫x ²/25 dx, 0,2/5

E(x) = 2x³/75 , 0,2/5

E(x) = 2(2/5)³/75

E(x) = 16/9375

4 0
3 years ago
Activity: A flagpole has cracked 1.7 m from the ground and fallen over, as if hinged. The top of the flagpole hit the ground 3 m
Anastasy [175]

Answer:

Height of flagpole before it fell = 5.15 m

Step-by-step explanation:

Given:

Height of remain pole (p) = 1.7 m

Distance from base (b) = 3 m

Find:

Height of flagpole before it fell

Computation:

Using Pythagorean theorem

H = √ p² + b

Length of broken part = √ 1.7² + 3²

Length of broken part = √ 2.89 + 9

Length of broken part = √ 11.89

Length of broken part = 3.45 (Approx)

Height of flagpole before it fell = Length of broken part + Height of remain pole

Height of flagpole before it fell = 3.45 + 1.7

Height of flagpole before it fell = 5.15 m

8 0
3 years ago
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