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spayn [35]
2 years ago
6

A spotlight is aimed at an angle of 50° up

Mathematics
1 answer:
Sophie [7]2 years ago
3 0
I don’t know the answer
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Can anybody help me with this bellwork?
aev [14]
First, find the amount of time is taken to travel 1 mile :
time taken to travel 1 mile = 20/6 = 3 ⅓ mins
hence, time taken to travel 9 miles = 3 ⅓ x 9 = 30 mins

time taken to travel 10 miles = 3 ⅓ x 10 = 33 ⅓ mins
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3 years ago
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find de number or ways you arrange 5 purple flowers, 3 yellow flowers, 2 red flowers, and 1 white flower in row.
Marta_Voda [28]

Answer: 30 I think

Step-by-step explanation:

5 0
3 years ago
Equivalent fractions
lyudmila [28]

Answer:

number 1. 5/4  number 2.  7/4 number 3. 1/5

Step-by-step explanation                                                                                           It's the shaped parts. Hope this helps : )

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2 years ago
Someone pls help me if u do tysm i have khan academy’s all due tmr ahh
Alexxandr [17]

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4 0
3 years ago
Problem 3.3.9 • (a) Starting on day 1, you buy one lottery ticket each day. Each ticket is a winner with probability 0.1. Find t
strojnjashka [21]

Answer:

a) The probability mass function of K = P(K=k) = \binom{k-1}{4}0.1^{4}*0.9^{k-5} ; k =5,6,...

b)

c)

Step-by-step explanation:

a) Let p be  the probability of winning each ticket be = 0.1

Then q which is the probability of failing each ticket  = 1 - p = 1  - 0.1 = 0.9

Assume X represents the  number of failure preceding the 5th success in x + 5 trials.

The last trial must be success whose probability is p = 0.1 and in the remaining (x + r- 1) ( x+ 4 ) trials we must have have (4) successes whose probability is given by:

\binom{x+r-1}{r-1}*p^{r-1}*q^{x} = \binom{x+4}{4}0.1^{4}*0.9^{x} ; x =0, 1, .........

Then, the probability distribution of random variable X is

P(X=x) = \binom{x+4}{4}0.1^{4}*0.9^{x} ; x =0, 1, .........

where;

X represents the  negative binomial random variable.

K= X + 5 = number of ticket buy up to and including fifth winning ticket.

Since K =X+5 this signifies that  X = K-5

as X takes value 0, 1 ,2,...

K takes value 5, 6 ,...

Therefore:

The probability mass function of K = P(K=k) = \binom{k-1}{4}0.1^{4}*0.9^{k-5} ; k =5,6,...

b)

Let p represent the probability of getting a tail on a flip of the coin

Thus p = 0.5 since it is a fair coin

where L = number of flips of the coin including 33rd occurrence of  tails

Thus; the negative binomial distribution of L can be illustrated as:

P(X=x) = \binom{x-1}{r-1}(1-p)^{x-r}p^r

where

X= L

r = 33  &

p = 0.5

Since we are looking at the 33rd success; L is likely to be : L = 33,34,35...

Thus; the PMF of L = P(L=l) = \binom{l-1}{33-1}(1-0.5)^{l}(0.5)^{33} \\ \\ \\  \mathbf{P(L=l) = \binom{l-1}{33-1}(0.5)^{l} }

c)  

Given that:

Let  M be the random variable which represents  the number of tickets need to be bought to get the first success,

also success probability is 0.01.

Therefore, M ~ Geo(0.01).

Thus, the PMF of M is given by:

P(M = m) = (1-0.01)^{m-1} * 0.01 ,  \ \  \ since \ \ \ (m = 1,2,3,4,....)

P(M=m) = (0.99)^{m-1} * 0.01 , m = 1,2,3,4,....

4 0
3 years ago
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