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Dima020 [189]
2 years ago
10

What is the scale factor from Figure A to Figure B?

Mathematics
1 answer:
JulijaS [17]2 years ago
5 0

Answer:

  • 1/3

Step-by-step explanation:

The figure B is the scaled copy of the figure A.

The both figures are similar to each other.

<u>The scale factor is the ratio of corresponding sides of figure B to A:</u>

  • k = 3/9 = 1/3

or

  • k = 1.2/3.6 = 1/3
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3 years ago
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Mai and Tyler work on the equation 2/5 b + 1 = -11 together. Mai’s solution is b= -25 and Tyler’s is b = -28 Here is their work.
alisha [4.7K]

Answer:

<em>I disagree with both solutions. The value of b that will make the expression correct is -30</em>

Step-by-step explanation:

Given the equation solved my Mai and Tyler expressed as:

2/5 b + 1 = -11

We are to check the veracity of the solutions;

2/5 b + 1 = -11

Subtract 1 from both sides of the expression

2/5 b + 1 -1 = -11-1

2/5 b = -12

Cross multiply

2b = -12 * 5

2b = -60

Divide both sides by 2

2b/2 = -60/2

b = -30

<em>Since the solution b = -25 and -28 does not tally with the gotten solution, I disagree with the both solutions</em>

6 0
3 years ago
Help me please<br> i only have 30 minutes until this is due
balu736 [363]

Answer:

4. sin 40= 0.7451131604793488384

5. cos 40= -0.666938061652261888

6. tan 50= -0.2719006119976306688

7. tan 40= -1.1172149309238962176

8. sin A= -0.5882162892399388672

9. cos A= -0.8087036521945458688

10. tan A= 0.7273570332515760128

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3 years ago
There are 28 students in a class. 13 of the students are boys. Two students from the class are chosen at random. a) If the first
Luba_88 [7]

Answer:

Step-by-step explanation:

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2 years ago
g A certain financial services company uses surveys of adults age 18 and older to determine if personal financial fitness is cha
leva [86]

Answer:

Null hypothesis:p_{1} = p_{2}    

Alternative hypothesis:p_{1} \neq p_{2}  

z=\frac{0.410-0.35}{\sqrt{0.385(1-0.385)(\frac{1}{1000}+\frac{1}{700})}}=2.502    

p_v =2*P(Z>2.502)= 0.0123    

Comparing the p value with the significance level assumed \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to to reject the null hypothesis, and we can say that the proportion analyzed is significantly different between the two groups at 5% of significance.    

Step-by-step explanation:

Data given and notation    

X_{1}=410 represent the number of people indicating that their financial security was more than fair for the recent year

X_{2}=245 represent the number of people indicating that their financial security was more than fair for the year before

n_{1}=1000 sample 1 selected  

n_{2}=700 sample 2 selected  

p_{1}=\frac{410}{1000}=0.410 represent the proportion estimated of indicating that their financial security was more than fair this year

p_{2}=\frac{245}{700}=0.35 represent the proportion estimated of indicating that their financial security was more than fair the year before

\hat p represent the pooled estimate of p

z would represent the statistic (variable of interest)    

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

\alpha significance level given  

Concepts and formulas to use    

We need to conduct a hypothesis in order to check if is there is a difference between the two proportions, the system of hypothesis would be:    

Null hypothesis:p_{1} = p_{2}    

Alternative hypothesis:p_{1} \neq p_{2}    

We need to apply a z test to compare proportions, and the statistic is given by:    

z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}   (1)  

Where \hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{410+245}{1000+700}=0.385  

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.    

Calculate the statistic  

Replacing in formula (1) the values obtained we got this:    

z=\frac{0.410-0.35}{\sqrt{0.385(1-0.385)(\frac{1}{1000}+\frac{1}{700})}}=2.502    

Statistical decision  

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

p_v =2*P(Z>2.502)= 0.0123    

Comparing the p value with the significance level assumed \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to to reject the null hypothesis, and we can say that the proportion analyzed is significantly different between the two groups at 5% of significance.    

7 0
3 years ago
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