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castortr0y [4]
4 years ago
9

For less than than an number x is eight

Mathematics
1 answer:
Lapatulllka [165]4 years ago
5 0
If for is four then x-4=8
thus would make x 12
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Write a polynomial function in standard form with zeros at 0, 1, and 2.
Nikitich [7]

Answer:

the first one is correct

6 0
3 years ago
Let​ T: set of real numbers R Superscript nℝnright arrow→set of real numbers R Superscript mℝm be a linear​ transformation, and
Klio2033 [76]

Answer:

\{T(v_1), T(v_2), T(v_3)\} is linearly dependent set.

Step-by-step explanation:

Given:  \{v_1,v_2,v_3\} is a linearly dependent set in set of real numbers R

To show: the set \{T(v_1), T(v_2), T(v_3)\} is linearly dependent.

Solution:

If \{v_1,v_2,v_3,...,v_n\} is a set of linearly dependent vectors then there exists atleast one k_i:i=1,2,3,...,n such that k_1v_1+k_2v_2+k_3v_3+...+k_nv_n=0

Consider k_1T(v_1)+k_2T(v_2)+k_3T(v_3)=0

A linear transformation T: U→V satisfies the following properties:

1. T(u_1+u_2)=T(u_1)+T(u_2)

2. T(au)=aT(u)

Here, u,u_1,u_2∈ U

As T is a linear transformation,

k_1T(v_1)+k_2T(v_2)+k_3T(v_3)=0\\T(k_1v_1)+T(k_2v_2)+T(k_3v_3)=0\\T(k_1v_1+k_2v_2+k_3v_3)=0\\

As \{v_1,v_2,v_3\} is a linearly dependent set,

k_1v_1+k_2v_2+k_3v_3=0 for some k_i\neq 0:i=1,2,3

So, for some k_i\neq 0:i=1,2,3

k_1T(v_1)+k_2T(v_2)+k_3T(v_3)=0

Therefore, set \{T(v_1), T(v_2), T(v_3)\} is linearly dependent.

6 0
4 years ago
29x97 broken apart <br> need help now <br> Factors willl mark brainlyest
IgorC [24]

I don't know what you mean by "Broken apart," but I'll do my best...

20 x 90 = 1800

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1800 + 63 = 1863

I hope this helped you! :)

6 0
2 years ago
Determine whether the following statement regarding the hypothesistest for two population proportions is true or false.
murzikaleks [220]

Answer:

False

Step-by-step explanation:

Let p1 be the population proportion for the first population

and p2 be the population proportion for the second population

Then

H_{0} p1  = p2

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Test statistic can be found usin the equation:

z=\frac{p1-p2}{\sqrt{{p*(1-p)*(\frac{1}{n1} +\frac{1}{n2}) }}} where

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As |p1-p2| gets smaller, the value of the <em>test statistic</em> gets smaller. Thus the probability of its being extreme gets smaller. This means its p-value gets higher.

As the<em> p-value</em> gets higher, the null hypothesis is less likely be rejected.

7 0
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zaharov [31]
1/2

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BUT since we are going from a large triangle to a small triangle, our answer needs to be less then one, making our answer 1/2
3 0
3 years ago
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