A rectangle has an area of 48

1 answer:

6 0

Answer: i didnt understand the answer choice but the correct answer is 48/12=width. width=4

Step-by-step explanation:
as we know the area of a rectangle is area=length X width
so we plug in 48 as area and 12 as length

48=12 x width
48/12=width(divide both sides by 12)

You might be interested in

One half of one one hundredth is

ella [17]

Answer:

1/2 of 1/100 = 1/200

Step-by-step explanation:

5 0

3 years ago

1/7 divided by 4 equals

gulaghasi [49]

1/ 28 is the answer.
= 1/7 ÷4
= 1/7 ÷ 4/1
=1/7 X 1/4
Multiply top by top, bottom by bottom
=1/28

If you do not know how to do it:
First make both sides into a fraction ( Whole numbers always have a 1 as a denominator).
Then you reciprocate (Flip) the second fraction.
Finally you multiply top by top, bottom by bottom
You get the answer!<span />

6 0

3 years ago

If a fair coin is flipped 15 times, what is the probability that there are more heads than tails?

ludmilkaskok [199]

Answer:

The probability that there are more heads than tails is equal to $\dfrac{1}{2}$ .

Step-by-step explanation:

Since the number of flips is an odd number, there can't be an equal number of heads and tails. In other words, there are either

more tails than heads, or,
more heads than tails.

Let the event that there are more heads than tails be $A$ . $\lnot A$ (i.e., not A) denotes that there are more tails than heads. Either one of these two cases must happen. As a result, $P(A) + P(\lnot A) = 1$ .

Additionally, since this coin is fair, the probability of getting a head is equal to the probability of getting a tail on each toss. That implies that (for example)

the probability of getting 7 heads out of 15 tosses will be the same as
the probability of getting 7 tails out of 15 tosses.

Due to this symmetry,

the probability of getting more heads than tails (A is true) is equal to
the probability of getting more tails than heads (A is not true.)

In other words $P(A) = P(\lnot A)$ .

Combining the two equations:

$\left\{\begin{aligned}&P(A) + P(\lnot A) = 1 \cr &P(A) = P(\lnot A)\end{aligned}\right.$ ,

$P(A) = P(\lnot A) = \dfrac{1}{2}$ .

In other words, the probability that there are more heads than tails is equal to $\dfrac{1}{2}$ .

This conclusion can be verified using the cumulative probability function for binomial distributions with $\dfrac{1}{2}$ as the probability of success.

$\begin{aligned}P(A) =& P(n \ge 8) \cr =& \sum \limits_{i = 8}^{15} {15 \choose i} (0.5)^{i} (0.5)^{15 - i}\cr =& \sum \limits_{i = 8}^{15} {15 \choose i} (0.5)^{15}\cr =& (0.5)^{15} \left({15 \choose 8} + {15 \choose 9} + \cdots + {15 \choose 15}\right) \cr =& (0.5)^{15} \left({15 \choose (15 - 8)} + {15 \choose (15 - 9)} + \cdots + {15 \choose (15 - 15)} \right) \cr =& (0.5)^{15} \left({15 \choose 7} + {15 \choose 6} + \cdots + {15 \choose 0}\right)\end{aligned}$