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4 years ago

8

Match sticks are used to make the pattern shown to make these 4squares ,you need 13match sticks how many squares in this pattern

could you make with 55 matchsticks

1 answer:

nata0808 [166]4 years ago

8 0

13 + 4 = 17

17 - 55 =38

38 is your answer

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A manufacturing company produces valves in various sizes and shapes. One particular valve plate is supposed to have a tensile st

maxonik [38]

Answer:

$z=\frac{5.0611-5}{\frac{0.2803}{\sqrt{42}}}=1.413$

$p_v =2*P(z>1.413)=0.158$

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the average tensile strength is different from 5lbs/mm at 10% of signficance

Step-by-step explanation:

Data given and notation

$\bar X=5.0611$ represent the sample mean

$\sigma=0.2803$ represent the population standard deviation for the sample

$n=42$ sample size

$\mu_o =5$ represent the value that we want to test

$\alpha=0.1$ 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 true mean is different from 5, the system of hypothesis would be:

Null hypothesis: $\mu = 5$

Alternative hypothesis: $\mu \neq 5$

If we analyze the size for the sample is > 30 but and we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-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:

$z=\frac{5.0611-5}{\frac{0.2803}{\sqrt{42}}}=1.413$

P-value

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

$p_v =2*P(z>1.413)=0.158$

Conclusion

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the average tensile strength is different from 5lbs/mm at 10% of signficance

8 0

4 years ago

Which is a unit rate?<br> A. 1/7 <br> B. 48/4<br> C. 34/1 <br> D. 40/5

BartSMP [9]

So,

Normally a unit rate will be x per 1y. Example: 60 miles/1hr.

Therefore, the option with the denominator as 1 would be the unit rate.

The correct option is C.

4 0

4 years ago

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jok3333 [9.3K]

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3 years ago

A source of information randomly generates symbols from a four letter alphabet {w, x, y, z }. The probability of each symbol is

koban [17]

The expected length of code for one encoded symbol is

$\displaystyle\sum_{\alpha\in\{w,x,y,z\}}p_\alpha\ell_\alpha$

where $p_\alpha$ is the probability of picking the letter $\alpha$ , and $\ell_\alpha$ is the length of code needed to encode $\alpha$ . $p_\alpha$ is given to us, and we have

$\begin{cases}\ell_w=1\\\ell_x=2\\\ell_y=\ell_z=3\end{cases}$

so that we expect a contribution of

$\dfrac12+\dfrac24+\dfrac{2\cdot3}8=\dfrac{11}8=1.375$

bits to the code per encoded letter. For a string of length $n$ , we would then expect $E[L]=1.375n$ .

By definition of variance, we have

$\mathrm{Var}[L]=E\left[(L-E[L])^2\right]=E[L^2]-E[L]^2$

For a string consisting of one letter, we have

$\displaystyle\sum_{\alpha\in\{w,x,y,z\}}p_\alpha{\ell_\alpha}^2=\dfrac12+\dfrac{2^2}4+\dfrac{2\cdot3^2}8=\dfrac{15}4$

so that the variance for the length such a string is

$\dfrac{15}4-\left(\dfrac{11}8\right)^2=\dfrac{119}{64}\approx1.859$

"squared" bits per encoded letter. For a string of length $n$ , we would get $\mathrm{Var}[L]=1.859n$ .

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3 years ago

Please help Linear equations

svlad2 [7]

Answer:

2 and 4

Step-by-step explanation:

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2 years ago