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3 years ago

Please help me find the slope of the line which one would it be ? :) and why

Mathematics

1 answer:

slava [35]3 years ago

8 0

I think C. It’s going to the right 2 and up 5.

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The Rocky Mountain News (January 24, 1994) indicated that the 20-year mean snowfall in the Denver/Boulder region is 28.76 inches

ycow [4]

Answer:

The p-value of the test is 0.0007 < 0.05, indicating that the the snowfall for the 1993-1994 winters was higher than the previous 20-year average.

Step-by-step explanation:

20-year mean snowfall in the Denver/Boulder region is 28.76 inches. Test if the snowfall for the 1993-1994 winters has higher than the previous 20-year average.

At the null hypothesis, we test if the average was the same, that is, of 28.76 inches. So

$H_0: \mu = 28.76$

At the alternate hypothesis, we test if the average incresaed, that is, it was higher than 28.76 inches. So

$H_1: \mu > 28.76$

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample.

28.76 is tested at the null hypothesis:

This means that $\mu = 28.76$

Standard deviation of 7.5 inches. However, for the winter of 1993-1994, the average snowfall for a sample of 32 different locations was 33 inches.

This means that $\sigma = 7.5, X = 33, n = 32$ .

Value of the test statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{33 - 28.76}{\frac{7.5}{\sqrt{32}}}$

$z = 3.2$

P-value of the test and decision:

The p-value of the test is the probability of finding a sample mean above 33, which is 1 subtracted by the p-value of z = 3.2. In this question, we consider the standard level $\alpha = 0.05$ .

Looking at the z-table, z = 3.2 has a p-value of 0.9993.

1 - 0.9993 = 0.0007

The p-value of the test is 0.0007 < 0.05, indicating that the the snowfall for the 1993-1994 winters was higher than the previous 20-year average.

5 0

3 years ago

PLEASE HELP!!!!! HAVING TROUBLE <br><br> (see attached images)

qaws [65]

Answer:

The Proof for

Part C , Qs 9 and Qs 10 is below.

Step-by-step explanation:

PART C .

Given:

AD || BC ,

AE ≅ EC

To Prove:

ΔAED ≅ ΔCEB

Proof:

Statement Reason

1. AD || BC 1. Given

2. ∠A ≅ ∠C 2. Alternate Angles Theorem as AD || BC

3. ∠AED ≅ ∠CEB 3. Vertical Opposite Angle Theorem.

4. AE ≅ EC 4. Given

5. ΔAED ≅ ΔCEB 5. By A-S-A congruence test....Proved

Qs 9)

Given:

AB ≅ BC ,

∠ABD ≅ ∠CBD

To Prove:

∠A ≅ ∠C

Proof:

Statement Reason

1. AB ≅ BC 1. Given

2. ∠ABD ≅ ∠CBD 2. Given

3. BD ≅ BD 3. Reflexive Property

4. ΔABD ≅ ΔCBD 4. By S-A-S congruence test

5. ∠A ≅ ∠C 5. Corresponding parts of congruent Triangles Proved.

Qs 10)

Given:

∠MCI ≅ ∠AIC

MC ≅ AI

To Prove:

ΔMCI ≅ ΔAIC

Proof:

Statement Reason

1. ∠MCI ≅ ∠AIC 1. Given

2. MC ≅ AI 2. Given

3. CI ≅ CI 3. Reflexive Property

4. ΔMCI ≅ ΔAIC 4. By S-A-S congruence test

4 0

3 years ago

Add 10 to z, then double the result<br> Do not simplify any part of the expression.

dybincka [34]

2(z+10)

2z+20

Answer:

2z+20

4 0

3 years ago

It says factor each expression 49x-28?

NARA [144]

Answer:

7(7x -4)

Step-by-step explanation:

49x-28

We can factor a 7 from each term

49x = 7*7x

28 = 7*4

7(7x -4)

3 0

4 years ago

Define the double factorial of n, denoted n!!, as follows:n!!={1⋅3⋅5⋅⋅⋅⋅(n−2)⋅n} if n is odd{2⋅4⋅6⋅⋅⋅⋅(n−2)⋅n} if n is evenand (

tekilochka [14]

Answer:

Radius of convergence of power series is $\lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}=\frac{1}{108}$

Step-by-step explanation:

Given that:

n!! = 1⋅3⋅5⋅⋅⋅⋅(n−2)⋅n n is odd

n!! = 2⋅4⋅6⋅⋅⋅⋅(n−2)⋅n n is even

(-1)!! = 0!! = 1

We have to find the radius of convergence of power series:

$\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}](8x+6)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}]2^{n}(4x+3)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}](x+\frac{3}{4})^{n}\\$

Power series centered at x = a is:

$\sum_{n=1}^{\infty}c_{n}(x-a)^{n}$

$\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}](8x+6)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}]2^{n}(4x+3)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}4^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}](x+\frac{3}{4})^{n}\\$

$a_{n}=[\frac{8^{n}4^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}]\\\\a_{n+1}=[\frac{8^{n+1}4^{n+1}n!(3(n+1)+3)!(2(n+1))!!}{[(n+1+9)!]^{3}(4(n+1)+3)!!}]\\\\a_{n+1}=[\frac{8^{n+1}4^{n+1}(n+1)!(3n+6)!(2n+2)!!}{[(n+10)!]^{3}(4n+7)!!}]$

Applying the ratio test:

$\frac{a_{n}}{a_{n+1}}=\frac{[\frac{32^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}]}{[\frac{32^{n+1}(n+1)!(3n+6)!(2n+2)!!}{[(n+10)!]^{3}(4n+7)!!}]}$

$\frac{a_{n}}{a_{n+1}}=\frac{(n+10)^{3}(4n+7)(4n+5)}{32(n+1)(3n+4)(3n+5)(3n+6)+(2n+2)}$

Applying n → ∞

$\lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}= \lim_{n \to \infty}\frac{(n+10)^{3}(4n+7)(4n+5)}{32(n+1)(3n+4)(3n+5)(3n+6)+(2n+2)}$

The numerator as well denominator of $\frac{a_{n}}{a_{n+1}}$ are polynomials of fifth degree with leading coefficients:

$(1^{3})(4)(4)=16\\(32)(1)(3)(3)(3)(2)=1728\\ \lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}=\frac{16}{1728}=\frac{1}{108}$

4 0

3 years ago