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Lorico [155]
2 years ago
10

A 16.0-inch chord is drawn in a circle whose radius is 10.0 inches. what is the angular size of the minor arc of this chord? wha

t is the length of the arc, to the nearest tenth of an inch?
Mathematics
1 answer:
statuscvo [17]2 years ago
8 0
PART I
Angular size of the minor arc . 
Half of the chord an the radius makes a right angled triangle with the radius as the hypotenuse and half of the chord as one of the shorter side.
Therefore, using trigonometric ratio, sine = opp/hyp
                          sine θ = 8/10  where θ is half the minor angle
                                  θ  = 53.13
Therefore, the angular size of the minor arc will be 53.13 × 2  = 106.26°

PART II
The length of an arc is given by (θ/360 )× 2πr
where θ is the angle subtended by the arc to the center of the circle and r is the radius of the circle.
Therefore, length = (106.26/360) × 3.142 × 2×10
                            = 18.548 Inches
                                   
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Answer: 345.02

Step-by-step explanation:

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198 cents to 234 cents using a fraction in simplest form
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Answer:

11/13

Step-by-step explanation:

198/234

(198/18)/(234/18)

11/13

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This year, a hardware store has a profit of -$6.0 million. This profit is 3/4 of last year's profit. What is last year's profit?
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4.5Million dollars Bam Boom
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2 years ago
Last year, 46% of business owners gave a holiday gift to their employees. A survey of business owners indicated that 35% plan to
DiKsa [7]

Answer:

a) Number = 60 *0.35=21

b) Since is a left tailed test the p value would be:  

p_v =P(Z  

c) If we compare the p value obtained and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% the proportion of business owners providing holiday gifts had decreased from the 2008 level.  

We can use as smallest significance level 0.044 and we got the same conclusion.  

Step-by-step explanation:

Data given and notation  

n=60 represent the random sample taken

X represent the business owners plan to provide a holiday gift to their employees

\hat p=0.35 estimated proportion of business owners plan to provide a holiday gift to their employees

p_o=0.46 is the value that we want to test

\alpha=0.05 represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

Part a

On this case w ejust need to multiply the value of th sample size by the proportion given like this:

Number = 60 *0.35=21

2) Part b

We need to conduct a hypothesis in order to test the claim that the proportion of business owners providing holiday gifts had decreased from the 2008 level:  

Null hypothesis:p\geq 0.46  

Alternative hypothesis:p < 0.46  

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.35 -0.46}{\sqrt{\frac{0.46(1-0.46)}{60}}}=-1.710  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided \alpha=0.05. The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

p_v =P(Z  

Part c

If we compare the p value obtained and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% the proportion of business owners providing holiday gifts had decreased from the 2008 level.  

We can use as smallest significance level 0.044 and we got the same conclusion.  

8 0
2 years ago
A sports statistician was interested in the relationship between game attendance (in thousands) and the number of
Luda [366]

Answer:

The predicted number of wins for a team that has an attendance of 2,100 is 25.49.

Step-by-step explanation:

The regression  equation for the relationship between game attendance (in thousands) and the number of  wins for baseball teams is as follows:

\hat y = 4.9\cdot x + 15.2

Here,

<em>y</em> = number of wins

<em>x</em> = attendance (in thousands)

Compute the number of wins for a team that has an attendance of 2,100 as follows:

\hat y = 4.9\cdot x + 15.2

  =(4.9\times 2.1)+15.2\\=10.29+15.2\\=25.49

Thus, the predicted number of wins for a team that has an attendance of 2,100 is 25.49.

4 0
2 years ago
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