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Nataliya [291]
3 years ago
14

Mr. randall bought 4 shirts, which were on sale. the shirts were originally priced $20. the salres price of the shirts was $5 do

llars less than the original price. write and find the value of an expression for the total amount that mr. randall paid for the shirts. (was supposed to be math)
Mathematics
1 answer:
USPshnik [31]3 years ago
8 0

Answer:

The price Mr. Randall paid = 4 * (20-5)

Step-by-step explanation:

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Your friend claims that it is possible to draw a right triangle where the cosine of either angle (theta) is exactly the same val
avanturin [10]

Answer:

B. No

Cos \ \theta\neq Cos (90-\theta)\textdegree

Step-by-step explanation:

-A right angle triangle has two complimentary acute angles and one right angle.

-\theta is usually one of the acute angles and is equivalent to 90º minus it's complimentary acute angle.

-Complimentary angles add up to 90º.

#For complimentary angles:

Sin \ \theta=Cos \ (90-\theta)\textdegree\\\\Cos \ \theta=Sin(90-\theta)\textdegree\\\\\therefore Cos \ \theta\neq Cos (90-\theta)\textdegree

The two acute angles cannot have the same Cosine value.

Hence, she's not correct.

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3 years ago
Please explain how you got the answer!
ValentinkaMS [17]
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<em>Not </em>a definition of additional: SAME, same, and same.

We are given that EF is parallel to GH.
Remember, the definition of additional.

We wouldn't want the same piece of information, would we?

That's why we want a new piece of information:
Answer Choice B; FG parallel to EH.
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3 years ago
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Please help me solve the questions in the image above​
Makovka662 [10]

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Step-by-step explanation:

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3 years ago
Make a box and whiskers plot of the data. average temperature in Pittsburgh in June 79,70,73,84,83,88,75,72,76,68,68,79,84,74,75
aliya0001 [1]
The box-and-whisker plot is attached.

We first order the data from least to greatest:
67,68,68,69,70,72,73,74,75,75,76,78,79,79,80,83,84,85,85,88

Now we find the median.  Since there are 20 data values, the median is between 75 and 76:
(75+76)/2 = 151/2 = 75.5

Now we find the Lower Quartile (LQ).  We do this by finding the median of the lower half of data (from the median down).  There are 10 values here; the median is between 70 and 72:
(70+72)/2 = 142/2 = 71

Find the Upper Quartile (UQ).  Do this by finding the median of the upper half of data (from the median up).  There are 10 values here; the median is between 80 and 83:
(80+83)/2 = 163/2 = 81.5

The middle bar of the box is over the median.  The left hand side of the box is over LQ, and the right hand side of the box is over UQ.  Now we draw a whisker from the box to the highest value, 88, and from the box to the lowest value, 67.

7 0
3 years ago
To evaluate the effect of a treatment, a sample is obtained from a population, with a mean of u = 30, and the treatment is admin
Murrr4er [49]

Answer:

a) t=\frac{31.3-30}{\frac{3}{\sqrt{16}}}=1.733  

p_v =2*P(t_{15}>1.733)=0.104  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the we don't have a significant effect for the new treatment at 5% of significance.  

b) t=\frac{31.3-30}{\frac{3}{\sqrt{36}}}=2.6  

p_v =2*P(t_{15}>2.6)=0.0201  

If we compare the p value and the significance level given \alpha=0.05 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 we have a significant effect for the new treatment at 5% of significance.  

c) When we increase the sample size we increse the probability of rejection of the null hypothesis since the z score tend to increase when the sample size increase.

Step-by-step explanation:

Data given and notation  

Part a: If the sample consists of n=16 individuals, are the data sufficient to conclude that the treatment has a significant effect using a two-tailed test with alpha = 0.05.

\bar X=31.3 represent the sample mean  

s=3 represent the sample standard deviation  

n=16 sample size  

\mu_o =30 represent the value that we want to test  

\alpha=0.05 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 30, the system of hypothesis are :  

Null hypothesis:\mu = 30  

Alternative hypothesis:\mu \neq 30  

Since we don't know the population deviation, is better apply a z 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{31.3-30}{\frac{3}{\sqrt{16}}}=1.733  

P-value  

First w eneed to find the degrees of freedom given by:

df=n-1=16-1 =15

Since is a two-sided test the p value would given by:  

p_v =2*P(t_{15}>1.733)=0.104  

Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the we don't have a significant effect for the new treatment at 5% of significance.  

Part b: If the sample consists of n=36 individuals, are the data sufficient to conclude that the treatment has a significant effect using a two-tailed test with alpha = 0.05.

Calculate the statistic  

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

t=\frac{31.3-30}{\frac{3}{\sqrt{36}}}=2.6  

P-value  

First w eneed to find the degrees of freedom given by:

df=n-1=16-1 =15

Since is a two-sided test the p value would given by:  

p_v =2*P(t_{15}>2.6)=0.0201  

Conclusion  

If we compare the p value and the significance level given \alpha=0.05 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 we have a significant effect for the new treatment at 5% of significance.  

Part c; Comparing your answer for parts a and b, how does the size of the sample influence the outcome of a hypothesis test

When we increase the sample size we increse the probability of rejection of the null hypothesis since the z score tend to increase when the sample size increase.

5 0
3 years ago
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