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Brrunno [24]
3 years ago
12

The expression 2(l + w) is used to calculate the perimeter of a rectangle, where l is length and w is width. If the length is Fr

action 2 over 3 units and the width is Fraction 1 over 3 units, what is the perimeter of the rectangle in units?
Fraction 2 over 3 unit


1 unit


1 Fraction 2 over 3 units


2 units
Mathematics
1 answer:
bulgar [2K]3 years ago
5 0
E E E E E? If I’m right please let me know
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Enter the volume in cubic inches of the can round your anwser to the nearest hundredth
blsea [12.9K]

Answer:

87.92in³

Step-by-step explanation:

Volume of cylinder formula:

V = \pi r^{2} h

Plug in the values using 3.14 for pi:

V = 3.14(2)^2(7)

V = 87.92in³

5 0
3 years ago
What is the explicit formula for this sequence? -7,-3,1,5
uranmaximum [27]

Answer:

a_{n} = 4n - 11

Step-by-step explanation:

These are the terms of an arithmetic sequence with n th term

a_{n} = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

d = - 3 - (- 7) = 1 - (- 3) = 5 - 1 = 4

and a₁ = - 7, hence

a_{n} = - 7 + 4(n - 1) = - 7 + 4n - 4 = 4n - 11

6 0
3 years ago
El productor de dos números naturales es 176 y el primero es de 5 unidades menor que el segundo de que numero se trata
Ludmilka [50]

the rock and I have a great idea to make sure that the following is a process that is a process that you are writing to be a good idea for you

4 0
2 years ago
Which of the following describes the function f(x) = -x - 5] +27
Andru [333]

real sorry i cold not answer your question i am in a test right now myself

3 0
3 years ago
A well-known brokerage firm executive claimed that 50% of investors are currently confident of meeting their investment goals. A
VMariaS [17]

Answer:

p_v =P(Z  

If we compare the p value obtained and using the significance level given \alpha=0.005 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 adults that said they are confident of meeting their goals  is significantly lower than 0.5 or 50% .  

Step-by-step explanation:

1) Data given and notation

n=200 represent the random sample taken

X represent the people that said they are confident of meeting their goals

\hat p=0.44 estimated proportion of adults that said they are confident of meeting their goals

p_o=0.5 is the value that we want to test

\alpha=0.005 represent the significance level

Confidence=99.5% or 0.995

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 0.5.:  

Null hypothesis:p\geq 0.5  

Alternative hypothesis:p < 0.5  

When we conduct a proportion test we need to use the z statistic, 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{0.44 -0.5}{\sqrt{\frac{0.5(1-0.5)}{200}}}=-1.697  

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=0.005. The next step would be calculate the p value for this test.  

Since is left tailed test the p value would be:  

p_v =P(Z  

If we compare the p value obtained and using the significance level given \alpha=0.005 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 adults that said they are confident of meeting their goals  is significantly lower than 0.5 or 50% .  

4 0
4 years ago
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