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Crank
3 years ago
15

Can someone please answer. There is one problem. There's a picture. Thank you!!

Mathematics
1 answer:
strojnjashka [21]3 years ago
6 0
It's 12ft because you start dividing 452.39 by 4 and then by pi. Then you get 36=r squared. If you take the square root of both sides, you get 6=radius. The diameter will be 12 ft.
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Help me out please? tysvm!
julia-pushkina [17]

Answer:

Yes and b = why an no

Step-by-step explanation:

with thing and b

6 0
3 years ago
Read 2 more answers
How do you solve -2(x + 3y) = 18
sergiy2304 [10]

solving equation in terms of x we get x=-9-3y

solving equation in terms of y we get y=-3-1/3x

Step-by-step explanation:

We need to solve the equation: -2(x + 3y) = 18

Solving the equation:

-2(x + 3y) = 18\\-2x-6y=18

Simplifying and finding value of x:

-2x-6y=18\\-2x=18+6y\\-2x/-2=18/-2+6y/-2\\x=-9-3y

We can simplify and find value of y as well:

-2x-6y=18\\-6y=18+2x\\-6y/-6=18/-6+2x/-6\\y=-3-1/3x

So, solving equation in terms of x we get x=-9-3y

solving equation in terms of y we get y=-3-1/3x

Keywords: Solving Equations

Learn more about Solving Equations at:

  • brainly.com/question/1563227
  • brainly.com/question/2403985
  • brainly.com/question/11229113

#learnwithBrainly

5 0
3 years ago
To save money, a local charity organization wants to target its mailing requests for donations to individuals who are most suppo
mars1129 [50]

Answer:

The 80% confidence interval for difference between two means is (0.85, 1.55).

Step-by-step explanation:

The (1 - <em>α</em>) % confidence interval for difference between two means is:

CI=(\bar x_{1}-\bar x_{2})\pm t_{\alpha/2,(n_{1}+n_{2}-2)}\times SE_{\bar x_{1}-\bar x_{2}}

Given:

\bar x_{1}=M_{1}=6.1\\\bar x_{2}=M_{2}=4.9\\SE_{\bar x_{1}-\bar x_{2}}=0.25

Confidence level = 80%

t_{\alpha/2, (n_{1}+n_{2}-2)}=t_{0.20/2, (5+5-2)}=t_{0.10,8}=1.397

*Use a <em>t</em>-table for the critical value.

Compute the 80% confidence interval for difference between two means as follows:

CI=(6.1-4.9)\pm 1.397\times 0.25\\=1.2\pm 0.34925\\=(0.85075, 1.54925)\\\approx(0.85, 1.55)

Thus, the 80% confidence interval for difference between two means is (0.85, 1.55).

3 0
3 years ago
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
Perform the following computations.You may use13≈0.333333,34= 0.75 and100301= 0.332226.(i). Compute13+34by using five significan
White raven [17]

Answer:

a. 1.0833

Absolute Error = 0.416667

Relative Error = 1.250002

b. 0.0011070

Absolute Error = 0.0011070

Relative Error = 0.003321

Step-by-step explanation:

Given

1/3 = 0.333333

3/4 = 0.75

100/301 = 0.332226

a.

1/3 + 3/4

= 0.333333 + 0.75

= 1.083333

= 1.0833 ------ Approximated to 5 significant digits

Absolute Error = |Real Value - Estimated Value|

Relative Error = Absolute Error/Real Value

Assume 1/3 to be the real value and 3/4 to be the estimated value

Absolute Error = |0.333333 - 0.75|

Absolute Error = |-0.416667|

Absolute Error = 0.416667

Relative Error = 0.416667/0.333333

Relative Error = 1.250002

b.

1/3 - 100/301

= 0.333333 - 0.332226

= 0.001107

= 0.0011070 ----- Approximated to 5 significant digits

Assume 1/3 to be real value and 100/301 to be estimated value

Absolute Error = 0.333333 - 0.332226

Absolute Error = 0.0011070

Relative Error = 0.0011070/0.333333

Relative Error = 0.003321

Absolute and relative errors are approximation errors and they are due to the discrepancy between an exact value and some approximation to them.

The absolute error is the magnitude of the difference between the exact value and the approximation. The relative error is the absolute error divided by the magnitude of the exact value

4 0
3 years ago
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