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3 years ago

Keith has 4 dogs. I have 3 times as many dogs as keith. How many dogs do I have?

Mathematics

2 answers:

Darina [25.2K]3 years ago

8 0

Simple, Multiply 4 and 3 to get 12

4 x 3 = 12

Juli2301 [7.4K]3 years ago

6 0

You have 12 dogs. 4 X 3 = 12

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9741 divided by 221 in fraction form

babymother [125]

We want the GCD of 9741: 3 * 17 * 191. In order to get that...
9741 is divisible by 3 and equal to 3247 * 3 therefore we have 3 * 3247.
3247 is divisible by 17 and equal to 191 * 17 therefore we have 3 * 17 * 191, these are all prime numbers so no more factorization for this part.

Now we need to prime factorize the 221.
221 is divisible by 13 and equal to 17 * 13 therefore we have 13 * 17 which are all prime numbers meaning no more factorization.

Now we want to factor out 17 from the numerator/denominator

9741 = 17 * 573
221 = 17 * 13

In other words...(Same thing cancel the common factor of 17)
$\frac{17 * 573}{17 * 13} = \frac{573}{13}$

6 0

3 years ago

Given the following information about glucose levels (in milligrams of glucose per 100 milliliters of blood.) ( from 9.5)

Svetllana [295]

Answer:

A. 16.385

B. 9.532

C. (16.358)^2 is different from (9.532)^2

a. sigma squared of non pregnant women is GREATER THAN sigma squared of pregnant women

Step-by-step explanation:

<h3>A and B. Standard Deviation 's' of both columns:</h3>

The formula for the standard deviation is:

s = √(∑(x - μ)²/(n))

where,

∑ = is the sum function

x = a number from the set

μ = mean of the set

n = is the amount of the numbers in the set.

the column of non-pregnant women is

[73, 61, 104, 75, 85, 65, 62, 98, 92, 106]

first find the mean of this set:

mean = μ = (sum of all the numbers)/(amount of numbers)

μ = [73 + 61 + 104 +75 + 85 + 65 + 62 + 98 + 92 + 106]/(10)

μ = 821/10

μ = 82.1

similarly, for pregnant women the mean is

μ = 80.125

now, to find the standard deviation 's', first we need to find the variance 's²'. So, what you have to do is subtract each value in the column with the mean 'μ' and square the result.

for non-pregnant you will get:

[ 82.81, 445.21, 479.61, 50.41, 8.41, 292.41, 404.01, 252.81, 98.01, 571.21]

for pregnant you will get:

[ 66.015, 15.015, 97.515, 221.265, 199.51, 102.515, 1.265, 23.76]

finally just sum all the numbers of a column and divide with the amount of numbers in that column (remember that col1 has 10 numbers and col2 has 8 numbers). And you will get your variance for both pregnant and non-pregnant women:

for non-pregnant = 2684.89/10

non-pregnant = 268.4 (this is the variance and it is denoted by s²)

for standard deviation 's', just take the square root of the variance

$\sqrt{268.4}$ = 16.385.

similarly standard deviation of pregnant women can be found to be:

$\sqrt{90.859}$ = 9.532.

A. non-pregnant 'S' = 16.385

B. pregnant 'S' = 9.532

<h3>C. CLAIM:</h3>

you only have to show whether the variance of the above two columns are different or not.

And YES, the variances of the two column are indeed different, hence you make the CLAIM as written in question(C)

<h3>Multiple Choice:</h3>

here you need to show how different are the two values:

recall the variances:

Column1 = 268.4 (non-pregnant)

Column2 = 9.532 (pregnant)

now you know that the variance of non-pregnant is GREATER THAN the variance of pregnant

3 0

3 years ago

PLEASE HELP ME me me me

kykrilka [37]

Answer:

Andre has the correct answer. When simplified his answer is equivalent to the original equation.

7 0

4 years ago

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Please help, GodBless.

MAXImum [283]

Answer:

m = - 6 :)

Step-by-step explanation:

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3 years ago

Please help with this

pentagon [3]

Answer:

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3 years ago