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3 years ago

A couple plans to have two children. What is the probability that they have exactly one boy?

Mathematics

2 answers:

Licemer1 [7]3 years ago

7 0

Answer:

1/2

Step-by-step explanation:

Setler79 [48]3 years ago

7 0

Male and female having two children. The chance of having a 1 boy is 50%

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What is (m+1)(m+1) multiplied

Zarrin [17]

Apply the "FOIL" method here. (m+1)(m+1) = m^2 + 1m + 1m + 1, or m^2 + 2m + 1.

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3 years ago

Find the length of a side of a square if its área is 4/25 cm2

USPshnik [31]

Answer:

2/5cm or 0.4cm

Step-by-step explanation:

Area of square = $Side^{2}$

Thus, Side = $\sqrt{Area}$

= $\sqrt{4/25}$

= 2/5 cm = 0.4cm

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3 years ago

What is the domain and range of f(x)=2.5x when x goes up to 800

Mandarinka [93]

Answer:

Domain: all real numbers. Range: (-∞, 2000]

Step-by-step explanation:

f(x)=2.5x is a simple "ramp" function, a linear function and a polynomial. As such, its domain contains "all real numbers." That value 800 defines the largest value that this f(x) can have: f(800) = 2.5(800) = 2000.

Thus, the range is "all real numbers from -∞ through and including 2000."

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3 years ago

Cathy used a probability simulator to pull 3 colored marbles from a bag and flip a coin 50 times. The results are shown in the t

aliya0001 [1]

Answer:

blue marble 12/50. tails 20/50

Step-by-step explanation:

12/50×20/50 =240/2500= 12/125 =9.6%

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3 years ago

you drop a ball from a height of 98 feet. at the same time, your friend throws a ball upward. the polynomials represent the heig

Ostrovityanka [42]

a) $h_0 -u_y t$

b) See interpretation below

Step-by-step explanation:

The motion of both balls is a free-fall motion: it means that the ball is acted upon the force of gravity only.

Therefore, this means that the motion of the ball is a uniformly accelerated motion, with constant acceleration equal to the acceleration of gravity:

$g=32 ft/s^2$

in the downward direction.

For the ball dropped from the initial height of $h_0 = 98 ft$ , the height at time t is given by

$h(t) = h_0 -\frac{1}{2}gt^2$ (1)

The ball which is thrown upward from the ground instead is fired with an initial vertical velocity $u_y$ , and its starting height is zero, so its position at time t is given by

$h'(t)=u_y t - \frac{1}{2}gt^2$ (2)

Therefore, the polynomial that represents the distance between the two balls is:

$h(t)-h'(t)=h_0 - \frac{1}{2}gt^2 - (u_y t - \frac{1}{2}gt^2) = h_0 -u_y t$

Now we interpret this polynomial, which is:

$\Delta h(t) = h_0 -u_y t$

which represents the distance between the two balls at time t.

The interpretation of the two terms is the following:

- The constant term, $h_0$ , is the initial distance between the two balls, at time t=0 (in fact, the first ball is still at the top of the building, while the second ball is on the ground). For this problem, $h_0 = 98 ft$

- The coefficient of the linear term, $u_y$ , is the initial velocity of the second ball; this terms tells us that the distance between the two balls decreases every second by $u_y$ feet.

5 0

3 years ago