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1 year ago

10

Evaluate the following expression when x=-4 and y=4 x³-x /4y

Formula1" title="\frac{x}{y}" alt="\frac{x}{y}" align="absmiddle" class="latex-formula">

1 answer:

Simora [160]1 year ago

7 0

Answer: - 60

Step-by-step explanation:

$\frac{(-4)^3 - (-4)}{4(4)^-4/4} \\\\= -60$

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Help i need answers

k0ka [10]

Answer:

The last one if it says to add and the second one if it says to subt.

Hope this helps!!

Step-by-step explanation:

3 0

2 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.

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

$\begin{aligned}\phantom{P(A)} =& \sum \limits_{i = 0}^{7} {15 \choose i} (0.5)^{15}\cr =& P(n \le 7) \cr =& P(\lnot A)\end{aligned}$ .

6 0

3 years ago

Leviafan [203]

Total tickets sold = 417

2 fractions goes to students.
1 fraction goes to adults.

Total fractions of tickets
= 2 + 1
= 3 fractions

Since adults = 1 fraction
1 fraction = 417 ÷ 3
= 139 tickets

3 0

3 years ago

Read 2 more answers

I need help finding the value of x n y. thank you

rewona [7]

Step-by-step explanation:

Here with reference angle 45°

Hypotenuse ( h) = y

Perpendicular (p) = x

base (b) = 5√2

We know

tan 45° = p/b

1 = x / 5√2

Therefore x = 5√2

Now

h = √(5√2)^2 + 5√2)^2

= √ 50 +50

= √100

= 10

Hope it helped

4 0

2 years ago

manuel ate 1/3 of the crackers on a plate. his brother ate 1/4 of the crackers. there were 5 crackers left on the plate. how man

Neporo4naja [7]

Answer:

there would have been 12 crackers on the plate at the start

Step-by-step explanation:

So 1/3 +1/4 = 7/12

Therefore 5/12 left on the plate …ie 5.

So total to start with

5/5×12=12

Ans 12

5 0

2 years ago