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kakasveta [241]
3 years ago
9

How do I figure this out

Mathematics
2 answers:
Karolina [17]3 years ago
8 0

Answer:

D

Step-by-step explanation:

hope it help correct me if wrong

nikitadnepr [17]3 years ago
6 0

Answer:

D

Step-by-step explanation:

yeah-ya....................... right?

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The slope is undefined

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Let the following sample of 8 observations be drawn from a normal population with unknown mean and standard deviation: 22, 18, 1
BigorU [14]

Answer:

a:  Sample mean: 20, Sample standard deviation: 1.732

b:  19.13 <µ < 20.87

c:  18.84< µ < 21.16

d:  as the confidence level increases, the interval becomes wider

Step-by-step explanation:

a:  Sample mean is found by adding up all the individual values of the sample, then dividing by the total number of values in the sample.  

Sample standard deviation is the square root of the variance

See the first attached photo for the calculations of these values.

We need to create a 80% confidence interval for the population.  Since n < 30, we will use a t-value with degree of freedom of 7 (the degree of freedom is always one less than the sample size.  

Look for the column on the t-distribution chart that has "area in two tails" of 0.20 (80%), and row 7 (degree of freedom)

The t-value is 1.415

See the second attached photo for the construction of the confidence interval

We need to create a 90% confidence interval for the population.  Since n < 30, we will use a t-value with degree of freedom of 7 (the degree of freedom is always one less than the sample size.  

Look for the column on the t-distribution chart that has "area in two tails" of 0.10 (90%), and row 7 (degree of freedom)

The t-value is 1.895

See the third attached photo for the construction of the confidence interval

3 0
3 years ago
Factor the trinomial. Factor out the GCF first.<br> 5x^2y - 30xy^2 - 35y^3
anzhelika [568]

Answer: 5x^2y - 30xy^2 - 35y^3=5y(x+y)(x-7y)

Step-by-step explanation:

GCF = Greatest common factor.

For example: 6 is the GCF of 12 and 18.

The given expression: 5x^2y - 30xy^2 - 35y^3

It can be written as

5x^2y - (5\times6)x( y\times y) - (5\times7)(y^2\times y)

Taking out common factor  5y , we get

5y(x^2-6xy-7y^2)\\\\ =5y(x^2+xy-7xy-7y^2)\  \ [\because-6xy=xy-7xy]\\\\=5y(x(x+y)-7y(x+y))\\\\= 5y((x+y)(x-7y))\\\\=5y(x+y)(x-7y)

Hence, 5x^2y - 30xy^2 - 35y^3=5y(x+y)(x-7y)

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Find the distance between points S(6,−3)
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.....................

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