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MA_775_DIABLO [31]
3 years ago
15

The circumference of a circle is like the ________ of a figure.

Mathematics
2 answers:
adelina 88 [10]3 years ago
8 0

Answer:

The answer would be 'B. perimeter'.

Nina [5.8K]3 years ago
5 0
The guy above me is right.
You might be interested in
Select the correct answer.
iris [78.8K]

The future value of $1,000 invested at 8% compounded semiannually for five years is \bold{\$ 1,480}

<u>Solution:</u>

\bold{A = P (1 + i )^{n}} ----------- equation 1

A = future value  

P= principal amount  

i = interest rate

n = number of times money is compounded  

P = 1000

i = 8 %

\mathrm{n} = \text { compounding period } \times \text {number of years}

(Compounding period for semi annually = 2)

\mathrm{n} = \text { compounding period } \times \text {number of years}

Dividing “i” by compounding period

i = \frac{8 \%}{2} = 0.04

Solving for future value using equation 1

\begin{array}{l}{A = 1000(1 + 0.04)^{10}} \\\\ {=1000 (1.04)^{10}}\end{array}

= 1480.2

\approx 1,480 \$

3 0
3 years ago
If gross pay is $1500, what will the net pay be after 23% total deduction?
Ierofanga [76]

Answer:2000

Step-by-step explanation:I think

6 0
3 years ago
If sin theta = sqrt(2/2), which could not be the value of theta?
Nataliya [291]

The solution would be like this for this specific problem:

sin(θ°) = √(2)/2 

θ° = 360°n + sin⁻¹(√(2)/2) and θ° = 360°n + 180° − sin⁻¹(√(2)/2)
θ° = 360°n + 45° and θ° = 360°n + 135° where n∈ℤ 

360°*0 + 45° = 45° 
360°*0 + 135° = 135° 
360°*1 + 45° = 405° 

<span>sin(225°) = -√(2)/2

</span>225 has an angle where sin theta= -(sqrt2)/2 therefore, the value of theta cannot be 225 degrees.

4 0
3 years ago
Read 2 more answers
How do you simplify 1 48/99 to the smallest number
blagie [28]
First you need to turn this into an improper fraction. Then, using your improper fraction, you simplify
3 0
3 years ago
Read 2 more answers
A TV station claims that 38% of the 6:00 - 7:00 pm viewing audience watches its evening news program. A consumer group believes
Elena L [17]

Answer:

z=\frac{0.340 -0.38}{\sqrt{\frac{0.38(1-0.38)}{830}}}=-2.374  

p_v =P(Z  

So the p value obtained was a very low value 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% of significance the proportion of people that regularly watch the TV station’s news program is significantly less than 0.38 .  

Step-by-step explanation:

1) Data given and notation

n=830 represent the random sample taken

X=282 represent the people that regularly watch the TV station’s news program

\hat p=\frac{282}{830}=0.340 estimated proportion of people that regularly watch the TV station’s news program

p_o=0.38 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)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the proportion is less than 0.38:  

Null hypothesis:p\geq 0.38  

Alternative hypothesis:p < 0.38  

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.

3) Calculate the statistic  

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

z=\frac{0.340 -0.38}{\sqrt{\frac{0.38(1-0.38)}{830}}}=-2.374  

4) 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  

So the p value obtained was a very low value 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% of significance the proportion of people that regularly watch the TV station’s news program is significantly less than 0.38 .  

3 0
3 years ago
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