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Goshia [24]
3 years ago
15

The Fancy Feet Dog kennel has twice as many large breed dogs as small breed dogs. There are 10 large breed dogs. Write the equat

ion for the total number of dogs. Let d = the number of sogs board at the Fancy Feet Dog Kennel.
A. d = 2(10 + 10)
B. d = 10 + 1/2(10)
C. d = 1/2(10) + 2(10)
D. d = 1/2(10 + 10)
Mathematics
1 answer:
yuradex [85]3 years ago
7 0
B because if there are 10 large dogs and that is two times the number of small dogs than there are half as many small dogs as big doge which would mean the equation is D=1/2(10)+10.
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Which statement describes a number with the smallest absolute value?
Snezhnost [94]
I believe C. The temperature is -23 degrees.

This is because, a negative (absolute value) becomes positive

Which would lead for |-23| to end up being |23|

Anytime a negative absolute value is placed, it becomes a positive and a positive stays positive.

So the smallest would be C then B later D and at last A
4 0
3 years ago
Tom and Sam shared equally one third of a chocolate bar. What fraction of the chocolate bar did each child get?
Savatey [412]
1/3 each , hope it helps

 
5 0
2 years ago
Read 2 more answers
The Precision Scientific Instrument Company manufactures thermometers that are supposed to give readings of 0°C at the freezing
PSYCHO15rus [73]

Answer:

-1.53 , 1.53

Step-by-step explanation:

It is basically asking you to find the percentiles. so for the bottom 6.3% you would type into your calculator 2nd, VARS, invNorm, area: .063, mean: 0, and Standard Deviation: 1 which gives you -1.53.

for the upper 6.3% you have to take 1-.063=.937 to get the upper percentile. now type into your calculator 2nd, VARS, invNorm, area: .937, mean: 0, and Standard Deviation: 1 which gives you 1.53.


I hope I helped :D

6 0
1 year ago
What is the mean absolute deviation of 42 31 1 1 2 4 5 6 7 3
FromTheMoon [43]

The mean absolute deviation is 10.52.

<h3>What is the mean absolute deviation?</h3>

The mean absolute deviation of a dataset is the average distance between each data point and the mean. It is a measure of variation of a data set.

MAD =1/n  ∑ l x - m(x) l

Where:

n = total numbers in the data set

x = number

m(x) - median

The first step is to determine the mean of the data set: (42 + 31 + 1 + 1 + 2 4+ 5 + 6 + 7 + 3) / 10 = 10.2

Find the absolute value of the difference between the number and the mean. Add the result | (42 - 10.2) + (31 - 10.2) + (1 _ 10.2)  + (1 - 10.2)  + (2 - 10.2)  + (4 - 10.2)  + (5 - 10.2)  + (6 - 10.2)  + (7 - 10.2)  + (3 - 10.2) = 105.2

Divide the sum by the total number in the dataset = 105.2 / 10 = 10.52

To learn more about mean absolute deviation, please check: brainly.com/question/27365177

#SPJ1

5 0
1 year ago
You know that in a specific population of rainbow trout 15% of the individuals carry intestinal parasites. Assume you obtain a r
kicyunya [14]

Answer:

a) 0.2316 = 23.16% probability that 0 carry intestinal parasites.

b) 0.4005 = 40.05% probability that at least two individuals carry intestinal parasites.

Step-by-step explanation:

For each trout, there are only two possible outcomes. Either they carry intestinal parasites, or they do not. Trouts are independent. This means that 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.

You know that in a specific population of rainbow trout 15% of the individuals carry intestinal parasites.

This means that p = 0.15

Assume you obtain a random sample of 9 individuals from this population:

This means that n = 9

a. Calculate the probability that __ (last digit of your ID number) carry intestinal parasites.

Last digit is 0, so:

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

P(X = 0) = C_{9,0}.(0.15)^{0}.(0.85)^{9} = 0.2316

0.2316 = 23.16% probability that 0 carry intestinal parasites.

b. Calculate the probability that at least two individuals carry intestinal parasites.

This is

P(X \geq 2) = 1 - P(X < 2)

In which

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

So

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

P(X = 0) = C_{9,0}.(0.15)^{0}.(0.85)^{9} = 0.2316

P(X = 1) = C_{9,1}.(0.15)^{1}.(0.85)^{8} = 0.3679

P(X < 2) = P(X = 0) + P(X = 1) = 0.2316 + 0.3679 = 0.5995

P(X \geq 2) = 1 - P(X < 2) = 1 - 0.5995 = 0.4005

0.4005 = 40.05% probability that at least two individuals carry intestinal parasites.

5 0
2 years ago
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