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leonid [27]
3 years ago
5

The mean absolute deviation tells you how spread out or how clustered around the mean a set of data is. This is the variation in

the data. To compare sets, a higher mean absolute deviation indicates that the data points are more spread out from the mean. A lower mean absolute deviation indicates dat
Mathematics
2 answers:
jonny [76]3 years ago
6 0

Answer:

hes right ↑

Step-by-step explanation:

tamaranim1 [39]3 years ago
4 0

Answer:

data points or values are nearer to the mean

Step-by-step explanation:

Given that mean absolute deviation tells you how to spread out or how clustered around the mean a set of data is.

Hence, in a comparison of sets, a higher mean absolute deviation indicates that the data points are more spread out from the mean.

On the other hand, a lower mean absolute deviation indicates that the data points are nearer to the mean than a data set with a higher mean absolute deviation

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Choice C

Step-by-step explanation:

4x+80=180 degrees

summation of all angles in a triangle is 180°

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3 years ago
5 divide 1/6 in simplest form
Savatey [412]

Answer:

30

Step-by-step explanation:

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N x 6/9 = 3/12<br><br> what is N?
rusak2 [61]
1

 

Simplify n\times \frac{6}{9}n×​9​​6​​ to \frac{2n}{3}​3​​2n​​

\frac{2n}{3}=\frac{3}{12}​3​​2n​​=​12​​3​​


2

 

Simplify \frac{3}{12}​12​​3​​ to \frac{1}{4}​4​​1​​

\frac{2n}{3}=\frac{1}{4}​3​​2n​​=​4​​1​​


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Multiply both sides by 33

2n=\frac{1}{4}\times 32n=​4​​1​​×3


4

 

Simplify \frac{1}{4}\times 3​4​​1​​×3 to \frac{3}{4}​4​​3​​

2n=\frac{3}{4}2n=​4​​3​​


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Divide both sides by 22

n=\frac{\frac{3}{4}}{2}n=​2​​​4​​3​​​​


6

 

Simplify \frac{\frac{3}{4}}{2}​2​​​4​​3​​​​ to \frac{3}{4\times 2}​4×2​​3​​

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7

 

Simplify 4\times 24×2 to 88

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3 years ago
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Paraphin [41]

Answer:

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Step-by-step explanation:

4 0
3 years ago
67% of owned dogs in the United States are spayed or neutered. Round your answers to four decimal places. If 48 owned dogs are r
oksian1 [2.3K]

Answer:

a) Probability that exactly 29 of them are spayed or neutered = 0.074

b) Probability that at most 33 of them are spayed or neutered = 0.66

c) Probability that at least 30 of them are spayed or neutered = 0.79

d) Probability that between 28 and 33 (including 28 and 33) of them are spayed or neutered = 0.574

Step-by-step explanation:

This is a binomial distribution question

probability of having a spayed or neutered dog, p  = 0.67

probability of having a dog that is not spayed or neutered, q = 1 - 0.67

q = 0.23

sample size, n = 48

According to binomial distribution formula:

P(X=r) = nCr p^r q^{n-r}

where nCr = \frac{n!}{(n-r)! r!}

a) Probability that exactly 29 of them are spayed or neutered

P(X= 29) = 48C29 * 0.67^{29} * 0.23^{19}\\P(X=29) = 0.074

b) Probability that at most 33 of them are spayed or neutered

P(X \leq 33) =1 -  P(X > 33)\\P(X \leq 33) =1 - 0.34\\P(X \leq 33) = 0.66

c) Probability that at least 30 of them are spayed or neutered

P(X \geq 30) = 1 - P(x < 30)\\P(X \geq 30) = 1 - 0.21\\P(X \geq 30) = 0.79

d) Probability that between 28 and 33 (including 28 and 33) of them are spayed or neutered.

P(28 \leq X \leq 33) = P(X=28) + P(X=29) + P(X=30) + P(X=31) + P(X=32) + P(X=33)\\P(28 \leq X \leq 33) = 0.053 + 0.074 + 0.095 + 0.112 + 0.121 + 0.119\\P(28 \leq X \leq 33) = 0.574

8 0
3 years ago
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