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Oksi-84 [34.3K]
3 years ago
10

What is the common difference of the arithmetic

Mathematics
1 answer:
PIT_PIT [208]3 years ago
8 0

Answer:

The sequence is not arithmetic

Step-by-step explanation:

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Most US adults have social ties with a large number of people, including friends, family, co-workers, and other acquaintances. I
xxTIMURxx [149]

Answer:

t=\frac{669-634}{\frac{732}{\sqrt{1700}}}=1.971    

p_v =2*P(t_{(1699)}>1.971)=0.049  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from 634 at 10% of signficance.

Step-by-step explanation:

Data given and notation  

\bar X=669 represent the sample mean

s=732 represent the sample standard deviation

n=1700 sample size  

\mu_o =68 represent the value that we want to test

\alpha=0. represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is different from 634, the system of hypothesis would be:  

Null hypothesis:\mu = 634  

Alternative hypothesis:\mu \neq 634  

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

t=\frac{669-634}{\frac{732}{\sqrt{1700}}}=1.971    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=1700-1=1699  

Since is a two sided test the p value would be:  

p_v =2*P(t_{(1699)}>1.971)=0.049  

Conclusion  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from 634 at 10% of signficance.

3 0
3 years ago
I need help it has to be correct
Bess [88]

Answer:

cosine = adjacent / hypotenuse

cosine 62º = JL / 17

JL = cosine 62º * 17

JL = .47 * 17

JL = 7.99

JL = 8.0

Step-by-step explanation:

5 0
3 years ago
Cos(0.5)+cos(0.5+2(3.14159265))+cos(0.5+4(3.14159265))
aleksandr82 [10.1K]

Answer:

2.63274768567

Step-by-step explanation:

First, note that 3.1415... is p, so 2(3.1415...) is 2pi and 4(3.1415...) is 4pi.

2pi and 4pi are full rotations, which will lead to the same cosine (i.e. cos(x) = cos (x+2pi) = cos(x+4pi) = ... = cos (x+2k*pi)).

So, the expression equals cos0.5 + cos 0.5 + cos0.5 = 3cos0.5 = 3(0.87758256189) = 2.63274768567

I hope this helps! :)

7 0
3 years ago
A metal conduit will be used as a pathway for wiring through a concrete block. The conduit is a 5 foot long rod with an outer di
kolbaska11 [484]

Answer:  586.59 cubic centimeters .

Step-by-step explanation:

As per given . we have

Inner diameter =  1.8 inches

⇒Inner radius :r = 0.9 in.  (radius is half of diameter)

= 0.9 x (2.54) = 2.286 cm  [∵ 1 in . = 2.54 cm]

Outer diameter = 2 inches

⇒Outer radius : R = 1 inch  = 2.54 cm

Height : h = 5 feet = 5 x(30.48) = 152.4 cm   [∵  1 foot = 30.48 cm]

The formula to find the volume of a hollow cylinder :

V=\pi(R^2-r^2)h , where R= outer radius , r= inner radius and h= height.

Now , the volume of metal in the conduit :

V=(3.14)(( 2.54)^2-(2.286)^2)(152.4)

V=(3.14)(6.4516-5.225796)(152.4)

V=(3.14)(1.225804)(152.4)

V=586.591342944\approx586.59\ cm^3

Hence, the volume of metal in the conduit is 586.59 cubic centimeters .

5 0
3 years ago
The ratio of the heights of triangles is 2:3. find out the ratio of their areas.
Vitek1552 [10]
The answer to the question is 2:3
4 0
3 years ago
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