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tensa zangetsu [6.8K]
3 years ago
7

Write the phase as an algebraic expression.

Mathematics
2 answers:
anzhelika [568]3 years ago
8 0
6p? Im not sure if i understand what youre saying
soldier1979 [14.2K]3 years ago
6 0

Answer:

6p is the answer

Step-by-step explanation:

6 time p= 6p

You might be interested in
I am really confused about this question for math.. any help would be really appreciated:
Elina [12.6K]

y= (\frac 1 4 )^x

A reflection about the x axis, about y=0, is the mapping (x',y')=(x,-y) so

y'= -y = - (\frac 1 4)^{x'}

A dilation of 2 is the mapping (x'',y'')=(2x', 2y')

So

x'=x''/2, y'=y''/2

y''/2= - (\frac 1 4)^{x''/2}

y'' =  - 2((\frac 1 4)^{1/2})^{x''}

y''= - 2(\frac 1 2)^{x''}

We can rewrite that without the primes and combine the powers of 2.

y =  - 2^{1-x}

Let's graph these and see if we're close,

Plot y= (1/4)^x, y= - (1/4)^{x}, y = - 2^{1-x}

6 0
3 years ago
The Wisconsin Dairy Association is interested in estimating the mean weekly consumption of milk for adults over the age of 18 in
Stolb23 [73]

Answer:

ME = 1.653 *\frac{7.9}{\sqrt{300}}= \pm 0.75

±0.75 ounce

Step-by-step explanation:

Assuming this complete question: The Wisconsin Dairy Association is interested in estimating the mean weekly consumption of milk for adults over the age of 18 in that state. To do this, they have selected a random sample of 300 people from the designated population. The following results were recorded: xbar=34.5 ounces, s=7.9 ounces Given this information, if the leaders wish to estimate the mean milk consumption with 90 percent confidence, what is the approximate margin of error in the estimate?

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X=34.5 represent the sample mean

\mu population mean (variable of interest)

s=7.9 represent the sample standard deviation

n=300 represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=300-1=299

Since the Confidence is 0.90 or 90%, the value of \alpha=0.1 and \alpha/2 =0.05, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,199)".And we see that t_{\alpha/2}=1.653

And the margin of error is given by:

ME = 1.653 *\frac{7.9}{\sqrt{300}}= \pm 0.75

±0.75 ounce

4 0
3 years ago
Read 2 more answers
Item 29<br> Divide.<br> 6.8÷4<br> 6.8÷4
Ket [755]

Answer:

1.7

Step-by-step explanation:

You can just use a calculator to make the work easier.

7 0
2 years ago
I need G PLZZZZZZZZZZZZZZZ
klasskru [66]
I can’t even read it
3 0
3 years ago
Find values for θ that make each statement true (show steps)
Ksju [112]

Answer:

Step-by-step explanation:

1. sinФ = cos 25

25 ° is in the between 0 and 90°

therefore it can simply represent

cosФ= sin (90-Ф) = sin (90 - 25) = sin 65

2. sin(Ф/3 + 10) = cos Ф

cos Ф = sin (90 -Ф)

sin(Ф/3 + 10) = cos Ф

sin(Ф/3 + 10) = cos Ф = sin(90-Ф)

Ф/3+10=90-Ф

10Ф+3/30 = 90-Ф

10Ф+3 = 30(90-Ф)

10Ф+3 = 2700-30Ф

10Ф+30Ф=2700-3

40Ф = 2697

Ф = 2697 / 40 = 67.425 ≅ 67.4°

4 0
3 years ago
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