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hichkok12 [17]
2 years ago
14

Please help me asap I am in class

Mathematics
1 answer:
lions [1.4K]2 years ago
4 0
2: 2/3; 3: 3/4; 4: 1/4; 5: 1/3; 6: 1/4; 7: 4/5; 8: 5/6-3/6=1/3
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Restaurateur Denny Valentine is evaluating two sites, Raymondville and Rosenberg, for his next restaurant. He wants to prove tha
Aleksandr-060686 [28]

Answer:

\bar X_1 represent the sample for the sample 1

\bar X_2 represent the sample for the sample 2

s_1 represent the deviation for the sample 1

s_2 represent the deviation for the sample 2  

n_1 = 81 represent the sample size for the group 1

n_2 = 81 represent the sample size for the group 2

And on this case if we create the system of hypothesis we have this:

Null hypothesis :\mu_1 -\mu_2 \leq 0

Alternative hypothesis: \mu_1 -\mu_2 >0

Step-by-step explanation:

Previous concepts

A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".  

The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".  

The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".  

On this caseHe wants to prove that Raymondville residents (population 1) dine out more often than Rosenberg residents (population 2). So this statement need to be on the alternative hypothesis. And the complement of the alternative hypothesis would be on the null hypothesis.  

Solution to the problem

We assume that we have the following statistics:

\bar X_1 represent the sample for the sample 1

\bar X_2 represent the sample for the sample 2

s_1 represent the deviation for the sample 1

s_2 represent the deviation for the sample 2  

n_1 = 81 represent the sample size for the group 1

n_2 = 81 represent the sample size for the group 2

And on this case if we create the system of hypothesis we have this:

Null hypothesis :\mu_1 -\mu_2 \leq 0

Alternative hypothesis: \mu_1 -\mu_2 >0

5 0
3 years ago
The table shows a linear function.
sergejj [24]

Answer:

\large\boxed{y=\dfrac{5}{3}x+9}

Step-by-step explanation:

The slope-intercept form of an equation of a line:

y=mx+b

<em>m</em><em> - slope</em>

<em>b</em><em> - y-intercept → (0, b)</em>

The formula of a slope:

m=\dfrac{y_2-y_1}{x_2-x_1}

From the table we have an y-intercept (0, 9) → b = 9.

Take other point from the table (-6, -1).

Calculate the slope:

m=\dfrac{-1-9}{-6-0}=\dfrac{-10}{-6}=\dfrac{5}{3}

Finally:

y=\dfrac{5}{3}x+9

7 0
3 years ago
The regular price of an item at a store is p dollars. The item is on sale for 20% off the regular price. Some of the expressions
Volgvan

Answer:

B

Step-by-step explanation:

5 0
3 years ago
Thompson and Thompson is a steel bolts manufacturing company. Their current steel bolts have a mean diameter of 144 millimeters,
valentinak56 [21]

Answer:

The probability that the sample mean would differ from the population mean by more than 2.6 mm is 0.0043.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from the population with replacement, then the distribution of the sample  means will be approximately normally distributed.

Then, the mean of the distribution of sample mean is given by,

\mu_{\bar x}=\mu

And the standard deviation of the distribution of sample mean  is given by,

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}

The information provided is:

<em>μ</em> = 144 mm

<em>σ</em> = 7 mm

<em>n</em> = 50.

Since <em>n</em> = 50 > 30, the Central limit theorem can be applied to approximate the sampling distribution of sample mean.

\bar X\sim N(\mu_{\bar x}=144, \sigma_{\bar x}^{2}=0.98)

Compute the probability that the sample mean would differ from the population mean by more than 2.6 mm as follows:

P(\bar X-\mu_{\bar x}>2.6)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}} >\frac{2.6}{\sqrt{0.98}})

                           =P(Z>2.63)\\=1-P(Z

*Use a <em>z</em>-table for the probability.

Thus, the probability that the sample mean would differ from the population mean by more than 2.6 mm is 0.0043.

8 0
3 years ago
In ΔABC, if m∠CAD = 29°, the m∠DAB is......
daser333 [38]

Answer:125

Step-by-step explanation:

6 0
3 years ago
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