What is the ninth term in the binomial expansion of (x – 2y)13? 329,472x5y8 –329,472x5y8

Suppose the weights of fish caught from pier 1 and pier 2 are normally distributed with equal population standard deviations. Th

madreJ [45]

Answer:

H₀: μ₁ = μ₂ vs, Hₐ: μ₁ > μ₂.

Step-by-step explanation:

A two-sample z-test can be performed to determine whether the claim made by the owner of pier 1 is correct or not.

It is provided that the weights of fish caught from pier 1 and pier 2 are normally distributed with equal population standard deviations.

The hypothesis to test whether the average weights of the fish in pier 1 is more than pier 2 is as follows:

H₀: The weights of fish in pier 1 is same as the weights of fish in pier 2, i.e. μ₁ = μ₂.

Hₐ: The weights of fish in pier 1 is greater than the weights of fish in pier 2, i.e. μ₁ > μ₂.

The significance level of the test is:

α = 0.05.

The test is defined as:

$z=\frac{(\bar x_{1}-\bar x_{2})-(\mu_{1}-\mu_{2})}{\sqrt{\frac{\sigma_{1}^{2}}{n_{1}}+\frac{\sigma_{2}^{2}}{n_{2}}}}$

The decision rule for the test is:

If the p-value of the test is less than the significance level of 0.05 then the null hypothesis will be rejected and vice-versa.

6 0

2 years ago

Which of the following are dimensionally consistent? (Choose all that apply.)(a) a=v / t+xv2 / 2(b) x=3vt(c) xa2=x2v / t4(d) x=v

Bumek [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

A

is dimensionally consistent

B

is not dimensionally consistent

C

is dimensionally consistent

D

is not dimensionally consistent

E

is not dimensionally consistent

F

is dimensionally consistent

G

is dimensionally consistent

H

is not dimensionally consistent

Step-by-step explanation:

From the question we are told that

The equation are

$A) \ \ a^3 = \frac{x^2 v}{t^5}$

$B) \ \ x = t$

$C \ \ \ v = \frac{x^2}{at^3}$

$D \ \ \ xa^2 = \frac{x^2v}{t^4}$

$E \ \ \ x = vt+ \frac{vt^2}{2}$

$F \ \ \ x = 3vt$

$G \ \ \ v = 5at$

$H \ \ \ a = \frac{v}{t} + \frac{xv^2}{2}$

Generally in dimension

x - length is represented as L

t - time is represented as T

m = mass is represented as M

Considering A

$a^3 = (\frac{L}{T^2} )^3 = L^3\cdot T^{-6}$

and $\frac{x^2v}{t^5 } = \frac{L^2 L T^{-1}}{T^5} = L^3 \cdot T^{-6}$

Hence

$a^3 = \frac{x^2 v}{t^5}$ is dimensionally consistent

Considering B

$x = L$

and

$t = T$

Hence

$x = t$ is not dimensionally consistent

Considering C

$v = LT^{-1}$

and

$\frac{x^2 }{at^3} = \frac{L^2}{LT^{-2} T^{3}} = LT^{-1}$

Hence

$v = \frac{x^2}{at^3}$ is dimensionally consistent

Considering D

$xa^2 = L(LT^{-2})^2 = L^3T^{-4}$

and

$\frac{x^2v}{t^4} = \frac{L^2(LT^{-1})}{ T^5} = L^3 T^{-5}$

Hence

$xa^2 = \frac{x^2v}{t^4}$ is not dimensionally consistent

Considering E

$x = L$

;

$vt = LT^{-1} T = L$

and

$\frac{vt^2}{2} = LT^{-1}T^{2} = LT$

Hence

$E \ \ \ x = vt+ \frac{vt^2}{2}$ is not dimensionally consistent

Considering F

$x = L$

and

$3vt = LT^{-1}T = L$ Note in dimensional analysis numbers are

not considered

Hence

$F \ \ \ x = 3vt$ is dimensionally consistent

Considering G

$v = LT^{-1}$

and

$at = LT^{-2}T = LT^{-1}$

Hence

$G \ \ \ v = 5at$ is dimensionally consistent

Considering H

$a = LT^{-2}$

,

$\frac{v}{t} = \frac{LT^{-1}}{T} = LT^{-2}$

and

$\frac{xv^2}{2} = L(LT^{-1})^2 = L^3T^{-2}$

Hence

$H \ \ \ a = \frac{v}{t} + \frac{xv^2}{2}$ is not dimensionally consistent

8 0

2 years ago

Which of the following best describes the expression 7x + 2y?

anastassius [24]

A.The sum of two products; there are two terms

5 0

2 years ago

Read 2 more answers

Suppose we have a group of 10 light users and 10 heavy users. what is the probability that exactly 3 of the 20 users are hiv pos

dmitriy555 [2]

Answer: The answer is 30%.

Step-by-step explanation: Given that there is a group of 10 light users and 10 heavy users. We are to find the probability that exactly 3 of the 20 users are HIV positive.

We have the following four possible cases -

(i) All 3 are light users.

(ii) 1 is a light user and 2 are heavy users.

(iii) 2 are light users and 1 is a heavy user.

(iv) All 3 are heavy user.

Since there is a 45% chance of a light user to be HIV positive and 55% chance of a heavy user to be HIV positive, so the required probability is given by

$p\\\\=\dfrac{3\times 0.45+1\times 0.45+2\times 0.55+2\times 0.45+1\times 0.55+3\times 0.55}{20}\\\\=\dfrac{6}{20}\\\\=0.3\\\\=30\%.$

Thus, the probability is 30.

4 0

2 years ago

The table shows data from Vicente's Form 1040. Gross income $75,000 Standard deduction $12,000 Taxable income $63,000 Total inco

Iteru [2.4K]

The answer is a+b+c = d it’s is equivalent to a and d

7 1

2 years ago

What is the ninth term in the binomial expansion of (x – 2y)13? 329,472x5y8 –329,472x5y8 –41,184x8y5 41,184x8y5