Find the area of the circle. Round to the nearest square meter. 1, 257m ^ 2 314m ^ 2 897m ^ 2 63m ^ 2

I got two more pls answer

polet [3.4K]

Answer:

33.) = 4

34.) = 2

35.) = 6

36.) = 6

37.) = 91

38.) = 35

39.) = 4

40.) = 28

41.) = 18

42.) = (-6)

Step-by-step explanation:

(10 + 6) ÷ [(9 – 5) × (5 – 4)²]

(16) ÷ (4) × (1)²

16/4 = 4

Thus, The answer is 4

(3² × 2) ÷ (5 + 8 – 5 + 1)

(9 × 2) ÷ (9)

18/9 = 2

Thus, The answer is 2

6² ÷ {2 + 5 – [9 – (6 + 2)]}

6² ÷ [2 + 5 – (9 – 8)]

36 ÷ [2 + 5 – (1)]

36 ÷ 7 – 1

36/6 = 6

Thus, The answer is 6

(3 × 3 × 2) ÷ (4 – 1)

18/3 = 6

Thus, The answer is 6

4 × 3 × 10 – 4² – (10 + 3)

120 – 16 – 13

120 – 29 = 91

Thus, The answer is 91

5 × [(22 – 1) × 2] ÷ (8 + 2 – 4)

5 × [(21) × 2] ÷ (6)

5 × 42 ÷ 6

5 × 7 = 35

Thus, The answer is 35

-2 × 4 ÷ [2 + (-4) – (-4)] – 4

-8 ÷ (2 – 4 + 4) – 4

-8 ÷ 2 – 4

-8/-2 = 8/2 = 4

Thus, The answer is 4

-4(-1 + 2) × 4 – 2 + (-1)

-4 × (1) × 4 – 2 – 1

(-4) × (-7) = 28

Thus, The answer is 28

-3 – (-3 × 3) + 6 × 2

-3 – (-9) + 12

-3 + 9 + 12

6 + 12 = 18

Thus, The answer is 18

6 – {3² + [3 × (-1)²]}

6 – [9 + (3 × 1)]

6 – (9 + 3)

6 – 12 = -6

Thus, The answer is (-6)

<u>-TheUnknownScientist 72</u>

7 0

3 years ago

Read 2 more answers

If we wanted to create a new 90% confidence interval from a different sample for the proportion of those with a two on one date

slava [35]

Answer: 271

Step-by-step explanation:

The formula we use to find the sample size is given by :-

$n=p(1-p)(\dfrac{z_{\alpha/2}}{E})^2$

, where $z_{\alpha/2}$ is the two-tailed z-value for significance level of $(\alpha)$

p = prior estimation of the proportion

E = Margin of error.

If prior estimation of the proportion is unknown, then we take p= 0.5 , the formula becomes

$n=0.5(1-0.5)(\dfrac{z_{\alpha/2}}{E})^2$

$n=0.25(\dfrac{z_{\alpha/2}}{E})^2$

Given : Margin of error : E= 0.05

Confidence level = 90%

Significance level $\alpha=1-0.90=0.10$

Using z-value table , Two-tailed z-value for significance level of $0.10$

$z_{\alpha/2}=1.645$

Then, the required sample size would be :

$n=0.25(\dfrac{1.645}{0.05})^2$

Simplify,

$n=270.6025\approx271$

Hence, the required minimum sample size =271

3 0

3 years ago

The cost C, in dollars, to have medical records transcribed is a function of time, t in hours. A physician's group spends C(t) =

Contact [7]

Answer:

B. $39550

Step-by-step explanation:

The expression for the function is not well-formatted, find the correct format in the solution bellow

Step one:

given data

We are told that the function of the cost is

$C(t) = 0.04t^3- 0.1t^2 + 5.5t$

instantaneous rate c(t) is the value obtained when we plug in 100 for t in the function above

substituting we have

$C(t) = 0.04(100)^3- 0.1(100)^2 + 5.5(100)\\\\\C(t) = 0.04*1000000- 0.1*10000 + 5.5*100\\\\C(t) = 40000-1000+550\\\\C(t) = 39550$

$39550

3 0

2 years ago

Plz help<br>which line segment is a reflection of AB in the x-axis

k0ka [10]

I think it would be (F,E)

5 0

3 years ago

Read 2 more answers

Two vertical poles of length 10 and 12 feet, respectively, stand 17 feet apart. A cable reaches from the top of one pole to the

Marat540 [252]

Answer:

$L=\sqrt{x^{2}+100}+\sqrt{x^{2}-34x+433} ft$

Step-by-step explanation:

let length of calble = L

distance from 10-foot pole = x ft

Moving from the top of the 10-foot pole to point x, we have

From the diagram that

$x^{2} +10^{2}=L_{1} ^{2} \\x^{2} + 100 = L_{1} ^{2}$

Take the square root of both sides

$\sqrt{x^{2}+100}=\sqrt{L_{1} ^{2}}\\ \sqrt{x^{2}+100}= L_{1}$

Moving from point x to the top of the 12-foot pole, we have

From the diagram that

$(17-x)^{2}+12^{2}=L_{2} ^{2} \\289-34x+x^{2} +144=L_{2} ^{2}\\x^{2} -34x+433=L_{2} ^{2}$

Take the square root of both sides

$\sqrt{x^{2}-34x+433}=\sqrt{L_{2} ^{2}} \\\sqrt{x^{2}-34x+433}=L_{2}$

Amount of cable used, L = L1 + L2

$L=\sqrt{x^{2}+100}+\sqrt{x^{2}-34x+433}$

3 0

3 years ago