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3 years ago

−5(p+3/5)=−4 solve for p

Mathematics

2 answers:

vovangra [49]3 years ago

7 0

Step-by-step explanation:

-5(p+3/5)= -4 distribute

-5p -3 = -4 addition

-5p = -1 division

answer = .2

defon3 years ago

3 0

Answer: p = 1/5

Step-by-step explanation:

-5(p + 3/5) = -4

Distribute

-5p - 3 = -4

Add 3 to both sides

-5p = -1

Divide by -5 on both sides

p = 1/5

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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.

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