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2 years ago

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Mathematics

1 answer:

Ostrovityanka [42]2 years ago

4 0

Answer:

See below

Step-by-step explanation:

P(A) = 3/8
P(B) = 2/7

(i)

P(A and B) = P(A) * P(B) = 3/8*2/7 = 3/28

(ii)

P(A or B) = P(A) + P(B) - P(A and B) = 3/8 + 2/7 - 3/28 = 1/56(21 + 16 - 6) = 31/56

(iii)

P(not A and not B) = P(not A) * P(not B) = (1 - 3/8)*(1 - 2/7) = 5/8*5/7 = 25/56

Another way:

P(not A and not B) = 1 - P(A or B) = 1 - 31/56 = 25/56

Outcomes with two fair coins: TT, TH, HT, HH

Outcomes with normal dice: 1 to 6

P(odd number) = 3/6 = 1/2

P(two heads and an even number) = 1/4*3/6 = 1/8

P(head and tail) = 1/2
P(prime) = 3/6 = 1/2 (primes are 2, 3 and 5)
P(head, tail and prime) = 1/2*1/2 = 1/4

P(two tails and odd number) = 1/4*3/6 = 1/8

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If the points (–2, –1), (–1, 0), (2, 3), and (–3, –2) are joined to form a straight line, at what point does the line intersect

Snowcat [4.5K]

Answer:

Step-by-step explanation:

Pick 2 points and use the slope formula

(–1, 0), (2, 3)

Slope =(y2−y1)/(x2−x1)

(3−0)(2−−1)

3/3

y=1x

Hope this helps! Plz award Brainliest : )

5 0

3 years ago

The formula T= 2pi sqrt(L/32) relates the time, T, in seconds for a pendulum with the length, L, in feet, to make one full swing

tester [92]

The length of pendulum is 2.485 feet

<h3>Solution:</h3>

Given that,

The formula T= 2pi sqrt(L/32) relates the time, T, in seconds for a pendulum with the length, L, in feet, to make one full swing back and forth

Therefore, the given formula is:

$T=2\pi \sqrt{\frac{L}{32} }$

We have to find the length of pendulum that makes one full swing in 1.75 seconds

So the modify the given equation to find "L"

$T=2\pi \sqrt{\frac{L}{32} }\\\\ \sqrt{\frac{L}{32} }=\frac{T}{2 \pi}\\\\\text{Taking square root on both sides }\\\\\frac{L}{32} = \frac{T^2}{4 \pi^2}\\\\L = \frac{T^2}{4 \pi^2} \times 32\\\\L = \frac{T^2}{\pi^2 } \times 8$

Substitute T = 1.75 seconds and $\pi = 3.14$

$L = \frac{1.75^2}{3.14 \times 3.14} \times 8\\\\L = \frac{3.0625}{9.8596} \times 8\\\\L = 2.485$

Thus length of pendulum is 2.485 feet approximately

7 0

3 years ago

Determine whether the rule represents an exponential function. Explain why or why not.

Alenkasestr [34]

Answer:

Yes.

Step-by-step explanation:

Exponential Parent Function: $f(x)= a(b^{x})$

If we write out the equation:

$y = 6(5)^{x}$

Whenever we have an exponent of x, it is a exponential function. The 5 is simply the base of the exponent and the 6 means we are vertically stretching the graph by a factor of 6.

5 0

3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

vodomira [7]

$12x^2-75=3(4x^2-25)=3(2^2x^2-5^2)=3((2x)^2-5^2)=3(2x+5)(2x-5)\\\\\text{Used:}\ a^2-b^2=(a+b)(a-b)\\\\Answer:\ 5$

8 0

3 years ago

Use undetermined coefficient to determine the solution of:y"-3y'+2y=2x+ex+2xex+4e3x

Kitty [74]

First check the characteristic solution: the characteristic equation for this DE is

r ² - 3r + 2 = (r - 2) (r - 1) = 0

with roots r = 2 and r = 1, so the characteristic solution is

y (char.) = C₁ exp(2x) + C₂ exp(x)

For the ansatz particular solution, we might first try

y (part.) = (ax + b) + (cx + d) exp(x) + e exp(3x)

where ax + b corresponds to the 2x term on the right side, (cx + d) exp(x) corresponds to (1 + 2x) exp(x), and e exp(3x) corresponds to 4 exp(3x).

However, exp(x) is already accounted for in the characteristic solution, we multiply the second group by x :

y (part.) = (ax + b) + (cx ² + dx) exp(x) + e exp(3x)

Now take the derivatives of y (part.), substitute them into the DE, and solve for the coefficients.

y' (part.) = a + (2cx + d) exp(x) + (cx ² + dx) exp(x) + 3e exp(3x)

… = a + (cx ² + (2c + d)x + d) exp(x) + 3e exp(3x)

y'' (part.) = (2cx + 2c + d) exp(x) + (cx ² + (2c + d)x + d) exp(x) + 9e exp(3x)

… = (cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)

Substituting every relevant expression and simplifying reduces the equation to

(cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)

… - 3 [a + (cx ² + (2c + d)x + d) exp(x) + 3e exp(3x)]

… +2 [(ax + b) + (cx ² + dx) exp(x) + e exp(3x)]

= 2x + (1 + 2x) exp(x) + 4 exp(3x)

… … …

2ax - 3a + 2b + (-2cx + 2c - d) exp(x) + 2e exp(3x)

= 2x + (1 + 2x) exp(x) + 4 exp(3x)

Then, equating coefficients of corresponding terms on both sides, we have the system of equations,

x : 2a = 2

1 : -3a + 2b = 0

exp(x) : 2c - d = 1

x exp(x) : -2c = 2

exp(3x) : 2e = 4

Solving the system gives

a = 1, b = 3/2, c = -1, d = -3, e = 2

Then the general solution to the DE is

y(x) = C₁ exp(2x) + C₂ exp(x) + x + 3/2 - (x ² + 3x) exp(x) + 2 exp(3x)

4 0

2 years ago

See image for question.​

See image for question.