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Fudgin [204]
2 years ago
7

The GCF of any two even numbers is always even. Determine whether the statement is true or false. If true, explain why. If false

, give a reason.
Mathematics
2 answers:
Alex_Xolod [135]2 years ago
7 0
<span>It is true that the GCF of any two even numbers is always even. GCF refers to the greatest common factor that two or more numbers have, and obviously it is going to be an even number. Take 18, 30, and 42 as your example - their GCF will be 6, which is an even number, as are those three numbers. GCF of odd numbers will be odd, and GCF of an odd and even number is also always even.</span>
atroni [7]2 years ago
5 0
<span>The question is asking us whether it is true that The GCF of any two even numbers is always even. GDF means the greatest common factor, for example the GCF of 24 and 32 is 8 - it's the biggest of their factors. The GCF is a multiple of all the (prime) common factors, and 2 is a common factor of two even numbers, so GCF will always be even - this is true!</span><span />
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18% of the the students at a school are in year 8 and 16% of the students are in year 9. If the school has 126 tera 8 students,
yarga [219]

Answer:

112

Step-by-step explanation:

The general form of all percentage questions is: the chunk= (some percentage) (of the whole), or c=p*w

We know that 18% of students are 8th grade and that is 126 students, so p = 0.18 and c=126 (126 students are 18% of the whole school)

126 = (0.18)w, divide both sides by 0.18

126/(0.18) = w = 700

9th graders are 16% of the school or 16% of 700 students

c = (0.16) 700 = 112 students

5 0
2 years ago
Use forward and backward difference approximations of O(h) and a centered difference approximation of O(h2) to estimate the firs
strojnjashka [21]

Answer:

Answer has explained below.

Step-by-step explanation:

Consider the function is:

F(x) = 25x3 – 6x2 +7x -88

Differentiate with respect to x, we get

F’(x) = 25. 3x2 – 6.2x + 7

       = 75x2 – 12x +7

At x = 2, we have

F (2) = 25(2)3 – 6(2)2 + 7(2)-88

        =102

And f’(2) = 75(2)2 – 12 (2) +7

               =283

Now, calculate forward divided difference as:

xi + 1 = xi + h

        =2 + 0.25

        =2.25

F (xi + 1) = f (2.25) = 25 (2.25)3 – 6(2.25)2 +7(2.25) -88

                            =182.21

f’(2) = f(2.25) – f(2) / 0.25 = 182.21 – 102 / 0.25

                                             = 320.84

Єt = 283 – 320.8 / 283 = -13.36%

Now calculate backward divided difference:

Xi-1 = xi –h = 2 – 0.25 = 1.75

F(xi-1)= f(1.8) = 25 . (1.8)3 -6 (1.8)2 + 7 (1.8) – 88

                       = 50.96

F’(2) = f(2) – f(1.8) / 0.25 = 102 – 50.96 / 0.25 = 204.16

Єt = 283 – 204.16 / 283 = 27.86%

Finally, centered divided difference is obtain by inserting f(xi+1) and f (xi-1):

F’(2) = f(2.25) – f(1.8) /2 x 0.25 = 320.84 -50.96 / 0.5 = 539.68

Єt = 283 – 539.68 / 283 = -90.7%

6 0
2 years ago
"An ordinance requiring that a smoke detector be installed in all previously constructed houses has been in effect in a particul
Galina-37 [17]

Answer:

a) Probability that the claim is rejected when the actual value of p is 0.8 = P(X ≤ 15) = 0.0173

b) Probability of not rejecting the claim when p = 0.7, P(X > 15) = 0.8106

when p = 0.6, P(X > 15) = 0.4246

c) Check Explanation

The error probabilities are evidently lower when 15 is replaced with 14 in the calculations.

Step-by-step explanation:

p is the true proportion of houses with smoke detectors and p = 0.80

The claim that 80% of houses have smoke detectors is rejected if in a sample of 25 houses, not more than 15 houses have smoke detectors.

If X is the number of homes with detectors among the 25 sampled

a) Probability that the claim is rejected when the actual value of p is 0.8 = P(X ≤ 15)

This is a binomial distribution problem

A binomial experiment is one in which the probability of success doesn't change with every run or number of trials (probability that each house has a detector is 0.80)

It usually consists of a number of runs/trials with only two possible outcomes, a success or a failure (we are sampling 25 houses with each of them either having or not having a detector)

The outcome of each trial/run of a binomial experiment is independent of one another.

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = 25 houses sampled

x = Number of successes required = less than or equal to 15

p = probability of success = probability that a house has smoke detectors = 0.80

q = probability of failure = probability that a house does NOT have smoke detectors = 1 - p = 1 - 0.80 = 0.20

P(X ≤ 15) = Sum of probabilities from P(X = 0) to P(X = 15) = 0.01733186954 = 0.01733

b) Probability of not rejecting the claim when p= 0.7 when p= 0.6

For us not to reject the claim, we need more than 15 houses with detectors, hence, th is probability = P(X > 15), but p = 0.7 and 0.6 respectively for this question.

n = total number of sample spaces = 25 houses sampled

x = Number of successes required = more than 15

p = probability that a house has smoke detectors = 0.70, then 0.60

q = probability of failure = probability that a house does NOT have smoke detectors = 1 - p = 1 - 0.70 = 0.30

And 1 - 0.60 = 0.40

P(X > 15) = sum of probabilities from P(X = 15) to P(X = 25)

When p = 0.70, P(X > 15) = 0.8105639765 = 0.8106

When p = 0.60, P(X > 15) = 0.42461701767 = 0.4246

c) How do the "error probabilities" of parts (a) and (b) change if the value 15 in the decision rule is replaced by 14.

The error probabilities include the probability of the claim being false.

When X = 15

(Error probability when p = 0.80) = 0.0173

when p = 0.70, error probability = P(X ≤ 15) = 1 - P(X > 15) = 1 - 0.8106 = 0.1894

when p = 0.60, error probability = 1 - 0.4246 = 0.5754

When X = 14

(Error probability when p = 0.80) = P(X ≤ 14) = 0.00555

when p = 0.70, error probability = P(X ≤ 14) = 0.0978

when p = 0.60, error probability = P(X ≤ 14) = 0.4142

The error probabilities are evidently lower when 15 is replaced with 14 in the calculations.

Hope this Helps!!!

6 0
2 years ago
What is the domain of validity for csctheta =start fraction 1 over sine theta end fraction? (1 point) all real numbers all real
dem82 [27]
We know the following relationship:

csc(\theta)=\frac{1}{sin(\theta)}

The domain of a function are the inputs of the function, that is, a function f is a relation that assigns to each element x in the set A exactly one element in the set B. The set A is the domain (or set of inputs) of the function and the set B contains the range (or set of outputs).Then applying this concept to our function csc(\theta) we can write its domain as follows:

1. D<span>omain of validity for csc(\theta):
</span>
D: \{\theta \in R/ sin(\theta) \neq 0 \} \\ In words: All \ \theta \ that \ are \ real \ values \ except \ those \ that \ makes \ sin(\theta)=0 
 
When:

sin(\theta)=0?

when:

\theta=..., -2\pi,-\pi,0,\pi,\2pi,3pi,...,k\pi

where k is an integer either positive or negative. That is:

sin(k\theta)=0 \ for \ k=...,-2,-1,0,1,2,3,...

To match this with the choices above, the answer is:

<span>"All real numbers except multiples of \pi"

</span>
2. which identity is not used in the proof of the identity 1+cot^{2}(\theta)=csc^{2}(\theta):

This identity can proved as follows:

sin^2{\theta}+cos^{2}(\theta)=1 \ Dividing \ by \ sin^{2}(\theta) \\ \\ \therefore \frac{sin^2{\theta}}{sin^{2}(\theta)}+\frac{cos^{2}(\theta)}{sin^{2}(\theta)}=\frac{1}{sin^{2}(\theta)} \\ \\ \therefore 1+cot^{2}(\theta)=csc^{2}(\theta)

The identity that is not used is as established in the statement above:

<span>"1 +cos squared theta over sin squared theta= csc2theta"

Written in mathematical language as follows:

</span>\frac{1+cos^{2}(\theta)}{sin^{2}(\theta)}=csc^{2}(\theta)<span>


</span>
4 0
2 years ago
Read 2 more answers
Which graph represents viable values for y = 2x, where x is the number of pounds of rice scooped and purchased from a bulk bin a
oee [108]

Answer:  Most Viable: On a coordinate plane, a straight line with a positive slope begins at point (0, 0), and ends at point (2.5, 5).  

Also possible, but only if someone scoops exact amounts (maybe pre-packaged for people who don't want to do their own scooping.): On a coordinate plane, blue diamonds appear at points (0, 0), (1, 2), (2, 4).

Step-by-step explanation:

The line beginning at (0,0) ending at (2.5, 5)  represents all the prices for any amount that the customer scoops.  for example, $5 for 2 1/2 pounds or $1 for 1/2 pound or  $2 for 1 pound would all be represented in the graphed line.

The graphs with negative values don't make sense. You can't scoop negative pounds!

<em>Again, good descriptions but difficult to sort out. Are you able to hit [enter] or [return] between options, or attach a screenshot?</em>

6 0
2 years ago
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