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julia-pushkina [17]
3 years ago
5

What is the constant in the algebraic expression to represent the patter below??

Mathematics
1 answer:
Hitman42 [59]3 years ago
8 0

Answer:

The constant in the algebraic expression to represent the pattern below is 4.

Step-by-step explanation:

Imortant Tip:

  • An Arithmetic sequence has a constant difference 'd' which can be determined by computing the differences of all the adjacent terms

Given the pattern

15,11,7,3,-1,...

computing the differences of all the adjacent terms

11-15=-4,\:\quad \:7-11=-4,\:\quad \:3-7=-4,\:\quad \:-1-3=-4

The difference between all the adjacent terms of the pattern is the same which is

d=-4

Hence, the given pattern represents the Arithmetic sequence.

Therefore, the constant in the algebraic expression to represent the pattern below is 4.

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Answer:

y =  45/m^2  −  49/4

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3 years ago
Suppose that we want to test the hypothesis with a significance level (alpha) of .05 that the climate has changed since industri
saw5 [17]

Answer: we reject the initial claim

(50. 3802, 51.6198)

Step-by-step explanation:

A)

The initial statement is that the mean temperature throughout history is 50°

This is the null hypothesis and it is denoted as H'

H': u = 50.

After taking a sample of 40 years, we realized that the mean temperature is 51°.

This is the alternative hypothesis, in contradiction to the alternative.

The alternative hypothesis is denoted as H1

H1: u > 50 ( upper tailed).

Sample size (n) = 40

Sample mean (x) = 51

Population standard deviation (σ) = 2

We use a z test to get the value of the test statistics.

We are using a z test because sample size is greater than 30 ( n = 40) and population standard deviation is given.

Z score = x - u/ (σ/√n)

Z score = 51 - 50 / (2/√40)

Z score = 1 / 0.3162

Z score = 3.16

Our level of significance is 5%, and the critical value at this level of significance is 1.645.

By comparing the z score relative to the critical value, since our z score is greater than 1.645, it implies that we are in the rejection region hence we reject the initial claim.

B)

We are to construct a 95% confidence interval for population mean temperature.

For upper limit

u = x + Zα/2 ×(σ/√n)

Where Zα/2 = 1.96 which is the critical value for a two tailed test at 5% level of significance.

For upper limit, we have that

u = 51 + 1.96 × (2/√40)

u = 51 + 1.96 (0.3162)

u = 51 + 0.6198

u = 51.6198.

For lower limit, we have that

u = 51 - 1.96 × (2/√40)

u = 51 - 1.96 (0.3162)

u = 51 - 0.6198

u = 50. 3802.

Hence the 95% confidence level for mean temperature is given as (50. 3802°, 51.6198°)

7 0
3 years ago
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Harlamova29_29 [7]

Answer:

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The answer will be 52. hope it helps
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