1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
pav-90 [236]
3 years ago
12

758% as a decimal to the thousands place is........

Mathematics
2 answers:
Whitepunk [10]3 years ago
8 0
758% as a decimal to the thousands place is 7.58.
fiasKO [112]3 years ago
6 0
7.580 is 758% as a decimal to the thousandths place. 
You might be interested in
The boiling temperature (in degrees Celsius) of platinum is 199 more than four times the boiling temperature $z$ (in degrees Cel
Zolol [24]
The correct answer is letter C
5 0
3 years ago
The place value of 9 0.496
Vesna [10]
The place value of 9 in 0.496 is hundredths place. 
8 0
3 years ago
Read 2 more answers
Select the factors of 10ax 5bx − 8ay − 4by.
Yakvenalex [24]
10ax + 5bx - 8ay - 4by = 5x(2a + b) - 4y(2a + b) = (5x - 4y)(2a + b)
option d is the correct answer.
4 0
3 years ago
Read 2 more answers
A veterinarian does research on the causes of enteroliths, stones that develop in the colon of horses. She suspects that feeding
dsp73

Answer:

On average, horses with and without enteroliths are fed the same amount of alfalfa per month

Step-by-step explanation:

Hello!

The veterinarian wants to research if eating alfalfa causes horses to have enteroliths.

The study variables are:

X₁: number of alfalfa flakes eaten over a month by a horse with enteroliths.

X₂: number of alfalfa flakes eaten over a month by a horse free of enteroliths.

She calculated a CI interval, with level 1 - α and the interval contained the cero.

With a significance level α, the hypotheses are:

H₀: μ₁ - μ₂ = 0

H₁: μ₁ - μ₂ ≠ 0

To decide whether you should reject or not the null hypothesis using a Confidence Interval is the following:

If the CI contains the number stated in the null hypothesis, the decision is to not reject the null hypothesis.

If the CI doesn't contain the number stated in the null hypothesis, then the decision is to reject the null hypothesis.

Since the interval contains the cero, if 1 - α and α are complementary, the decision is to reject the null hypothesis.

This means that at a level of significance of α, you can conclude that there is no difference between the population means of the numbers of alfalfa flakes eaten by horses with or without enteroliths.

I hope it helps!

7 0
3 years ago
a set v is given, together with definitions of addition and scalar multiplication. determine which properties of a vector space
agasfer [191]

The properties of a vector space are satisfied Properties 1,2, 5(a) and 5(c) are satisfied, the relaxation of the homes aren't legitimate are ifv = x ^ 2 1× v=1^ ×x ^ 2 = 1 #V

Property three does now no longer follow: Suppose that Property three is legitimate, shall we namev = a * x ^ 2 +bx +cthe neuter of V. Since v is the neuter, then O have to be constant with the aid of using the neuted, consequently 0 = O + v = (O  x ^ 2 + Ox + O) + (a × x ^ 2 + bx + c) = c × x ^ 2 + b ^ 2 + a

= 0 If O is the neuter, then it ought to restore x², but 0+ x² = (0x²+0x+zero) + (x²+0x+zero) = 1.This is a contradiction due to the fact x² isn't 1. We finish that V doesnt have a neuter vector. This additionally method that belongings four would not observe either. A set with out 0 cant

have additive inverse

Let r= v ×2x ^ 2 + v × 1x +v0 , w= w ×2x ^ 2 + w × 1x +w0 . We have that\\v+w= (vO + wO) ^  x^ 2 +(vl^ × wl)^  x+ ( v 2^ × w2)• w+v= (wO + vO) ^x^ 2 +(wl^ × vl)x+ ( w 2^ ×v2)

Since the sum of actual numbers is commutative, we finish that v + w = w + v Therefore, belongings 5(a) is valid.

Property 5(b) isn't valid: we are able to introduce

a counter example. we could use z = 1 thenv = x ^ 2 w = x ^ 2 + 1\\(v + w) + z = (x ^ 2 + 2) + 1 = 3x ^ 2 + 1

v + (w + z) = x ^ 2 + (2x ^ 2 + 1) = x ^ 2 + 3

Since 3x ^ 2 +1 ne x^ 2 +3. then the associativity rule doesnt hold.

(1+2)^ * (x^ 2 +x)=3^ * (x ^ 2 + x) = 3x + 3\\1^ × (x^ 2 +x)+2^ × (x ^ 2 + x) = (x + 1) + (2x + 2) = 3x ^ 2 + x ( ne 3x + 3 )\\(1^ ×2)^ ×(x^ 2 +x)=2^ × (x ^ 2 + x) = 2x + 2\\1^ × (2^ × (x ^ 2 + x) )=1^ × (2x+2)=2x^ 2 +2x( ne2x+2)

Property f doesnt observe because of the switch of variables. for instance, if v = x ^ 2 1 × v=1^ × x ^ 2 = 1 #V

Properties 1,2, 5(a) and 5(c) are satisfied, the relaxation of the homes arent legitimate.

Step-with the aid of using-step explanation:

Note that each sum and scalar multiplication entails in replacing the order from that most important coefficient with the impartial time period earlier than doing the same old sum/scalar multiplication.

Property three does now no longer follow: Suppose that Property three is legitimate, shall we name v = a × x ^ 2 +bx +c the neuter of V. Since v is the neuter, then O have to be constant with the aid of using the neuted, consequently0 = O + v = (O × x ^ 2 + Ox + O) + (a × x ^ 2 + bx + c) = c × x ^ 2 + b ^ 2 + a

= zero If O is the neuter, then it ought to restore x², but zero + x² = (0x²+0x+zero) + (x²+0x+zero) = 1.This is a contradiction due to the fact x² isn't 1. We finish that V doesnt have a neuter vector. This additionally method that belongings four would not observe either. A set with out 0 cant have additive inverse

Let r= v × 2x ^ 2 + v × 1x +v0 , w= w2x ^ 2 + w × 1x +w0 . \\We have thatv+w= (vO + wO) ^ x^ 2 +(vl^ wl)^x+ ( v 2^ w2)w+v= (wO + vO) ^ x^ 2 +(wl^ vl)x+ ( w 2^v2)

Since the sum of actual numbers is commutative, we finish that v + w = w + v Therefore, belongings 5(a) is valid.

Property 5(b) isn't valid: we are able to introduce

a counter example. we could usez = 1 then v = x ^ 2 w = x ^ 2 + 1(v + w) + z = (x ^ 2 + 2) + 1 = 3x ^ 2 + 1v + (w + z) = x ^ 2 + (2x ^ 2 + 1) = x ^ 2 + 3\\Since 3x ^ 2 +1 ne x^ 2 +3.then the associativity rule doesnt hold.

Note that each expressions are same because of the distributive rule of actual numbers. Also, you could be aware that his assets holds due to the fact in each instances we 'switch variables twice.

· (1+2)^ * (x^ 2 +x)=3^ * (x ^ 2 + x) = 3x + 31^ * (x^ 2 +x)+2^ * (x ^ 2 + x) = (x + 1) + (2x + 2) = 3x ^ 2 + x ( ne 3x + 3 )(1^ * 2)^ * (x^ 2 +x)=2^ * (x ^ 2 + x) = 2x + 21^ * (2^ * (x ^ 2 + x) )=1^ ×* (2x+2)=2x^ 2 +2x( ne2x+2)

Read more about polynomials :

brainly.com/question/2833285

#SPJ4

8 0
11 months ago
Other questions:
  • Is 5.43 greater or less than 5.432?
    13·1 answer
  • The area of a rectangle is
    5·2 answers
  • Perform the following Operations on fraction 1/5÷4/5=?​
    9·1 answer
  • Charles has $40 and spends 1/4 of it on some shirts. how much does Charles haven left? A. $25 B. $40 C. $30 D. $10
    8·1 answer
  • Music, in 2001, full-length cassettes represented 3.4% of total music sales. Between 2001 and 2006, the percent decreased by abo
    8·1 answer
  • Harry collected green balls in a bag. He drew two balls, of the five balls, out of the bags. Is this a random sample of the gree
    15·1 answer
  • What is the missing value in the image above? *
    12·1 answer
  • Please help with this/
    10·1 answer
  • Give possible values for the measures of angles A and Cif ABC is a right triangle.
    13·1 answer
  • I need help please I will give a bunch of points I need the answer or explanation please
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!