Gary mails 10^3 flyers in one week so how many did he mail

PLSSS HELP IF YOU TURLY KNOW THISS

leonid [27]

Answer:

$\frac{1}{5^{9} }$

Step-by-step explanation:

Using the rule of exponents

$a^{-m}$ = $\frac{1}{a^{m} }$ , then

$5^{-9}$ = $\frac{1}{5^{9} }$

5 0

2 years ago

Read 2 more answers

- Olivia skates twice as far as Stephanie. This sum of their distances is 96 miles per month. How many

kotegsom [21]

Answer:

B.64 miles

Step-by-step explanation:

Let Oliver's be x

Let Stephanie's be y

$x = 2y - - - (a) \\ x + y = 96 - - - (b)$

substitute for x in (b):

$2y + y = 96 \\ 3y = 96 \\ y = 32$

find x in (a):

$x = (2 \times 32) \\ x = 64$

5 0

3 years ago

Read 2 more answers

• Machine s is cleaned every 12 weeks. •Machine that is cleaned every 8 weeks. What is the fewest number of weeks that will pass

vfiekz [6]

Week 24,assuming that both machines are cleaned at week 0, because multiples of 12 are 12, 24,36. And multiples of 8 are 8, 16, 24. 24 is also the LCM.

7 0

3 years ago

4. Assume that the chances of a basketball player hitting a 3-pointer shot is 0.4 and the probability of hitting a free-throw is

vovikov84 [41]

Answer:

0.0334 = 3.34% probability that the player will make exactly 3 3-pointers and 5-free throws.

Step-by-step explanation:

For each 3-pointer shot, there are only two possible outcomes. Either the player makes it, or the player does not. The same is valid for free throws. This means that both the number of 3-pointers and free throws made are given by binomial distributions.

Since 3-pointers and free throws are independent, first we find the probability of making exactly 3 3-pointers out of 10, then the probability of making exactly 5 free throws out of 10, and then the probability that the player will make exactly 3 3-pointers and 5-free throws is the multiplication of these probabilities.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Probability of making 3 3-pointers out of 10:

The chances of a basketball player hitting a 3-pointer shot is 0.4, which means that $p = 0.4$ . So this is $P(X = 3)$ when $n = 10$ .

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 3) = C_{10,3}.(0.4)^{3}.(0.6)^{7} = 0.21499$

Probability of making 5 free throws out of 10:

The probability of hitting a free-throw is 0.65, which means that $p = 0.65$ . The probability is $P(X = 5)$ when $n = 10$ .

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 3) = C_{10,5}.(0.65)^{5}.(0.35)^{5} = 0.15357$

Calculate the probability that the player will make exactly 3 3-pointers and 5-free throws.

0.21499*0.15537 = 0.0334

0.0334 = 3.34% probability that the player will make exactly 3 3-pointers and 5-free throws.

5 0

3 years ago

Subtract the following polynomials, then place the answer in the proper location on the grid.

Grace [21]

(3m - 4n - 7) - (8m + n - 6)

3m - 8m = - 5m
-4n - n = -5n
-7 - (-6) = -7 + 6 = -1

-5m - 5n - 1 is your answer

hope this helps

8 0

3 years ago