When graphed, which function has a horizontal asymptote at 4? A. f(x)= 2(3^x)+4
B. f(x)=2x-4
C. f(x)=-3x+4
D f(x)= 3(2^x)-4

Answer:

Step-by-step explanation:

We are to find critical values for the test given

a) df =16: Alpha = 0.01  and right tailed

Critical value= 2.583

b) df = 23-1 = 22: alpha = 0.1 and left tailed

critical= -1.717

c)df=31-1 =30:  alpha =0.1:  two tailed

t =1.697

Critical values can be obtained from critical t tables.

Left tailed will have negative sign and right tailed positive

Final answer:

The critical values for these tests of a population standard deviation can be found via looking up a chi-square distribution table at the specified degrees of freedom and alpha level. For a two-tail test, the alpha value needs to be divided equally in the two tails.

Explanation:

To determine the critical values for these tests of a population standard deviation, we first need to understand the critical values for a chi-squared test. The chi-square test is used when the degrees of freedom and the level of significance (alpha) are known.

(a) A​ right-tailed test with 16 degrees of freedom at the alpha equals 0.01 level of significance: To find this critical value, we would check a chi-square distribution table at 16 degrees of freedom and alpha equals 0.01. The value we find is the critical value.

(b) A​ left-tailed test for a sample of size n equals 23 at the alpha equals 0.1 level of significance: Similarly, we would check the chi-square distribution table but this time at 22 degrees of freedom and alpha equals 0.1. Please note that degrees of freedom is calculated as n-1 which gives us 22 in this case.

(c) A​ two-tailed test for a sample of size n equals 31 at the alpha equals 0.1 level of significance: For a two-tailed test, we distribute the alpha equally in the two tails of the distribution. That means, we lookup chi-square distribution table for 30 degrees of freedom and alpha equals 0.05 to get our critical value.

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Answer:

210.33 cm^2

Step-by-step explanation:

We know that 6 equilateral triangles makes one hexagon.

Also, an equilateral triangle has all its sides equal.

If the tile of each side of the triangular tile measure 9 cm, then the height of the triangular tiles can be gotten using Pythagoras's Theorem.

The triangle formed by each tile can be split along its height, into two right angle triangles with base (adjacent) 4.5 cm and slant side (hypotenuse) of 9 cm. The height  (opposite) is calculated as,

From Pythagoras's theorem,

substituting, we have

81 = 20.25 +

= 81 - 20.25 = 60.75

opp = = 7.79 cm  this is the height of the right angle triangle, and also the height of the equilateral triangular tiles.

The area of a triangle =

where b is the base = 9 cm

h is the height = 7.79 cm

substituting, we have

area = x 9 x 7.79 = 35.055 cm^2

Area of the hexagon that will be formed = 6 x area of the triangular tiles

==> 6 x 35.055 cm^2 = 210.33 cm^2

Answer:

a) P(A) = 0,4607     or   P(A) = 46,07 %

b) P(B) = 0,7120   or 71,2 %

c) P(C) = 0,2055  or P(C) = 20,55 %

Step-by-step explanation:

We will use two concepts in solving this problem.

1.- The probability of an event (A) is for definition:

P(A) = Number of favorable events/ Total number of events FE/TE

2.- If A and B are complementary events ( the sum of them is equal to 1) then:

P(A) = 1 - P(B)

a) The total number of events is:

C ( 24,4) = 24! / 4! ( 24 - 4 )!    ⇒  C ( 24,4) = 24! / 4! * 20!

C ( 24,4) = 24*23*22*21*20! / 4! * 20!  

C ( 24,4) = 24*23*22*21/4*3*2

C ( 24,4) = 24*23*22*21/4*3*2    ⇒  C ( 24,4) =  10626

TE = 10626

Splitting the group of tanks in two 6 with h-v  and 24-6 (18) without h-v

we get that total number of favorable events is the product of:

FE = 6* C ( 18, 3)  = 6 * 18! / 3!*15!  =  18*17*16*15!/15!

FE =  4896

Then P(A) ( 1 tank in the sample contains h-v material is:

P(A) = 4896/10626

P(A) = 0,4607     or   P(A) = 46,07 %

b) P(B) will be the probability of at least 1 tank contains h-v

P(B) = 1 - P ( no one tank with h-v)

Again Total number of events is 10626

The total number of favorable events for the ocurrence of P is C (18,4)

FE = C (18,4) = 18! / 14!*4! = 18*17*16*15*14!/14!*4!

FE = 18*17*16*15/4*3*2  = 3060

Then P = 3060/10626

P = 0,2879

And the probability we are looking for is

P(B) = 1 - 0,2879

P(B) = 0,7120   or 71,2 %

c) We call P(C) the probability of finding exactly 1 tank with h-v and t-i

having 4 with t-i tanks is:

reasoning the same way but now having 4 with t-i (impurities) number of favorable events is:

FE = 6*4* C(14,2) = 24 * 14!/12!*2!

FE = 24* 14*13*12! / 12!*2

FE = 24*14*13/2    ⇒  FE = 2184

And again as the TE = 10626

P(C) = 2184/ 10626

P(C) = 0,2055  or P(C) = 20,55 %

Final answer:

These problems relate to the field of probability and specifically utilize the Hypergeometric Distribution. By plugging data into the appropriate formula, we can find the probabilities. For instance, the joint probability of independent events can be used to find the chance of exactly one tank having high viscosity and one tank having high impurities.

Explanation:

This question is asking us to solve probability problems. It specifically related to the Hypergeometric Distribution, which is used when we're interested in success/failure outcomes (in this case, tanks with high or acceptable viscosity), and when we're sampling without replacement.

For part a, we are looking for the probability of picking exactly one tank with high-viscosity. We would calculate this using the hypergeometric distribution formula:

P(X=k) = (C(K, k) * C(N-K, n-k)) / C(N, n)

where K is the total number of success states in the population (6 tanks with high viscosity), k is the number of success states in the sample (1 tank), N is the population size (24 tanks), and n is the number of samples (4 tanks). Plugging these numbers in, we can find the answer.

For part b, to find the probability that at least one tank in the sample contains high-viscosity material, we can either sum the probabilities P(X=1), P(X=2), P(X=3), and P(X=4), or find the complement of the probability that none of the tanks have high viscosity, i.e., 1- P(X=0).

For part c, with the addition of 4 tanks with high impurities, we can use the joint probability of independent events, which is the product of the probabilities of the two independent events. Here, the probability that exactly one tank has high viscosity and exactly one tank has high impurities would be the product of the two individual probabilities calculated in a similar manner.

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The compound interest is applied to the remaining balance in the account

each subsequent year.

• The amount that must be deposited is approximately $620.07

Reasons:

The given compound interest rate, r = 6.25% = 0.0625

The balance in the account after 2 years, A = $700

Required:

The required deposit, P, that gives the given account balance after 2 years.

Solution:

The following is the compound interest formula to use;

Where;

t = 2 years

We get;

Which gives;

• The amount that must be deposited to give $700 after 2 years is P ≈ $620.07

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Answer:

D. (2, -3/4)

Step-by-step explanation:

Using the substitution method:

-5x+4(1/8x-1)=-13

-5x+0.5x-4=-13

-4.5x/4.5=-9/4.5

-x=-2

x=2

You are supposed to replace 2 in the first equation now but as there is no other option with x value of 2 D is the answer.

If continued:

-5(2)+4y=-13

-10+4y=-13

4y=-3

y=-3/4