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Olin [163]
3 years ago
13

At the dog show, there are 4 times as many boxers as spaniels. If there are a total of 30 dogs, how many dogs are spaniels?

Mathematics
2 answers:
soldi70 [24.7K]3 years ago
5 0
There are 6 spaniel because 6 times 4 equals 24 and if you add 6 + 24 it equals 30 the amount of dogs at the show...hope it makes sense I messed up a little explaining
Scorpion4ik [409]3 years ago
5 0
If there are 6 spaniels and you multiply that by 4 it would come out to 24. 24 plus 6(the spaniels) then it would come out to 30
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Please help!!! I will mark best answer!
Kisachek [45]

Answer:

D

Step-by-step explanation:

7 0
2 years ago
In one town, the number of burglaries in a week has a poisson distribution with a mean of 1.9. find the probability that in a ra
tester [92]

Let X be the number of burglaries in a week. X follows Poisson distribution with mean of 1.9

We have to find the probability that in a randomly selected week the number of burglaries is at least three.

P(X ≥ 3 ) = P(X =3) + P(X=4) + P(X=5) + ........

= 1 - P(X < 3)

= 1 - [ P(X=2) + P(X=1) + P(X=0)]

The Poisson probability at X=k is given by

P(X=k) = \frac{e^{-mean} mean^{x}}{x!}

Using this formula probability of X=2,1,0 with mean = 1.9 is

P(X=2) = \frac{e^{-1.9} 1.9^{2}}{2!}

P(X=2) = \frac{0.1495 * 3.61}{2}

P(X=2) = 0.2698

P(X=1) = \frac{e^{-1.9} 1.9^{1}}{1!}

P(X=1) = \frac{0.1495 * 1.9}{1}

P(X=1) = 0.2841

P(X=0) = \frac{e^{-1.9} 1.9^{0}}{0!}

P(X=0) = \frac{0.1495 * 1}{1}

P(X=0) = 0.1495

The probability that at least three will become

P(X ≥ 3 ) = 1 - [ P(X=2) + P(X=1) + P(X=0)]

= 1 - [0.2698 + 0.2841 + 0.1495]

= 1 - 0.7034

P(X ≥ 3 ) = 0.2966

The probability that in a randomly selected week the number of burglaries is at least three is 0.2966

5 0
2 years ago
A small town had a population 960 people last year. The population grew to 1200 people this year. By what percentage did the pop
Rufina [12.5K]

Answer:

25%

Step-by-step explanation:

1200/960 = 1.25

So it grew by 25%

3 0
3 years ago
A particular telephone number is used to receive both voice calls and fax messages. Suppose that 25% of the incoming calls invol
bagirrra123 [75]

Answer:

a) 0.214 = 21.4% probability that at most 4 of the calls involve a fax message

b) 0.118 = 11.8% probability that exactly 4 of the calls involve a fax message

c) 0.904 = 90.4% probability that at least 4 of the calls involve a fax message

d) 0.786 = 78.6% probability that more than 4 of the calls involve a fax message

Step-by-step explanation:

For each call, there are only two possible outcomes. Either it involves a fax message, or it does not. The probability of a call involving a fax message is independent of other calls. So we use the binomial probability distribution to solve this question.

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.

25% of the incoming calls involve fax messages

This means that p = 0.25

25 incoming calls.

This means that n = 25

a. What is the probability that at most 4 of the calls involve a fax message?

P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4).

In which

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

P(X = 0) = C_{25,0}.(0.25)^{0}.(0.75)^{25} = 0.001

P(X = 1) = C_{25,1}.(0.25)^{1}.(0.75)^{24} = 0.006

P(X = 2) = C_{25,2}.(0.25)^{2}.(0.75)^{23} = 0.025

P(X = 3) = C_{25,3}.(0.25)^{3}.(0.75)^{22} = 0.064

P(X = 4) = C_{25,4}.(0.25)^{4}.(0.75)^{21} = 0.118

P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.001 + 0.006 + 0.025 + 0.064 + 0.118 = 0.214

0.214 = 21.4% probability that at most 4 of the calls involve a fax message

b. What is the probability that exactly 4 of the calls involve a fax message?

P(X = 4) = C_{25,4}.(0.25)^{4}.(0.75)^{21} = 0.118

0.118 = 11.8% probability that exactly 4 of the calls involve a fax message.

c. What is the probability that at least 4 of the calls involve a fax message?

Either less than 4 calls involve fax messages, or at least 4 do. The sum of the probabilities of these events is 1. So

P(X < 4) + P(X \geq 4) = 1

We want P(X \geq 4). Then

P(X \geq 4) = 1 - P(X < 4)

In which

P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

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

P(X = 0) = C_{25,0}.(0.25)^{0}.(0.75)^{25} = 0.001

P(X = 1) = C_{25,1}.(0.25)^{1}.(0.75)^{24} = 0.006

P(X = 2) = C_{25,2}.(0.25)^{2}.(0.75)^{23} = 0.025

P(X = 3) = C_{25,3}.(0.25)^{3}.(0.75)^{22} = 0.064

P(X

P(X \geq 4) = 1 - P(X < 4) = 1 - 0.096 = 0.904

0.904 = 90.4% probability that at least 4 of the calls involve a fax message.

d. What is the probability that more than 4 of the calls involve a fax message?

Very similar to c.

P(X \leq 4) + P(X > 4) = 1

From a), P(X \leq 4) = 0.214)

Then

P(X > 4) = 1 - 0.214 = 0.786

0.786 = 78.6% probability that more than 4 of the calls involve a fax message

8 0
2 years ago
You have a TDEE of
Gemiola [76]

Answer:

look down

Step-by-step explanation:

we need to find 10% and multiply it by 4 to get 40 percent

10 percent of 2200 is 220 because 2200 divided by 10 is 220 then we multiply this by 4 and we should get 880

3 0
2 years ago
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