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Vesnalui [34]
3 years ago
13

1. What was the second of two crucial events that increased U.S. concerns about the spread of

Mathematics
2 answers:
Deffense [45]3 years ago
6 0

Answer:

1.The berlin wall was built

2.Germany

3.1947

sineoko [7]3 years ago
4 0
1. the answer is D 
2. You are correct it is vietnam
3. the answer is 1947
You might be interested in
The number of undergraduates at Johns Hopkins University is approximately 2000, while the number at Ohio State University is app
Likurg_2 [28]

The remaining part of Question:

4) what can we conclude about sampling variability in the sample proportion calculated in the sample at John Hopkins as compared to that calculated in the sample at Ohio State.

5) The number of undergraduates at Johns Hopkins is approximately 2000 while the number at Ohio State is approximately 40000, suppose instead that at both schools, a simple random sample of about 3% of the undergraduates Will be taken.

Answer:

4) The sample proportion from Johns Hopkins will have about the same sampling variability as that from Ohio State

5) The sample proportion from John Hopkins will have more sampling variability than that from Ohio State

Step-by-step explanation:

Note: The sampling variability in the sample proportion decreases with increase in the sample size.

4) since the sample size at both Johns Hopkins and Ohio State is the same (i.e. n = 50), the sample variability of the sample proportion will be the same for both cases.

5) 3% of the population are selected for the observation in both cases.

At Johns Hopkins, sample size, n = 3% * 2000

n = 60

At Ohio State, sample size, n = 3% * 40000

n = 1200

Since sampling variability in the sample proportion decreases with increase in the sample size, the sampling variability in sample proportion will be higher at Johns Hopkins than at Ohio State.

5 0
3 years ago
The amount of a radioactive isotope decays in half every year the amount of the isotope can be modeled by f(x)=3461/2x and f(1)=
pychu [463]

Answer:

1 represents the number of years passed.

step -by-step explanation:

The amount of a radioactive isotope decays in half every year. The amount of the isotope can be modeled by f(x) = 346 (1/2)x and f(1) = 173

Here 1 represents the number of years that passed.

So 1 represents the number of years.

Hope this will helpful.

Thank you.

4 0
3 years ago
Field book of an land is given in the figure. It is divided into 4 plots . Plot I is a right triangle , plot II is an equilatera
Kazeer [188]

Answer:

Total area = 237.09 cm²

Step-by-step explanation:

Given question is incomplete; here is the complete question.

Field book of an agricultural land is given in the figure. It is divided into 4 plots. Plot I is a right triangle, plot II is an equilateral triangle, plot III is a rectangle and plot IV is a trapezium, Find the area of each plot and the total area of the field. ( use √3 =1.73)

From the figure attached,

Area of the right triangle I = \frac{1}{2}(\text{Base})\times (\text{Height})

Area of ΔADC = \frac{1}{2}(\text{CD})(\text{AD})

                        = \frac{1}{2}(\sqrt{(AC)^2-(AD)^2})(\text{AD})

                        = \frac{1}{2}(\sqrt{(13)^2-(19-7)^2} )(19-7)

                        = \frac{1}{2}(\sqrt{169-144})(12)

                        = \frac{1}{2}(5)(12)

                        = 30 cm²

Area of equilateral triangle II = \frac{\sqrt{3} }{4}(\text{Side})^2

Area of equilateral triangle II = \frac{\sqrt{3}}{4}(13)^2

                                                = \frac{(1.73)(169)}{4}

                                                = 73.0925

                                                ≈ 73.09 cm²

Area of rectangle III = Length × width

                                 = CF × CD

                                 = 7 × 5

                                 = 35 cm²

Area of trapezium EFGH = \frac{1}{2}(\text{EF}+\text{GH})(\text{FJ})

Since, GH = GJ + JK + KH

17 = \sqrt{9^{2}-x^{2}}+5+\sqrt{(15)^2-x^{2}}

12 = \sqrt{81-x^2}+\sqrt{225-x^2}

144 = (81 - x²) + (225 - x²) + 2\sqrt{(81-x^2)(225-x^2)}

144 - 306 = -2x² + 2\sqrt{(81-x^2)(225-x^2)}

-81 = -x² + \sqrt{(81-x^2)(225-x^2)}

(x² - 81)² = (81 - x²)(225 - x²)

x⁴ + 6561 - 162x² = 18225 - 306x² + x⁴

144x² - 11664 = 0

x² = 81

x = 9 cm

Now area of plot IV = \frac{1}{2}(5+17)(9)

                                = 99 cm²

Total Area of the land = 30 + 73.09 + 35 + 99

                                    = 237.09 cm²

7 0
3 years ago
Find the measure of angle E.
9966 [12]
Add all of the interior angles to 1260. Then you simplify for x. After that you plug in what you found for X in the E expression. 

4 0
3 years ago
ANYONE...........help​
KATRIN_1 [288]

Answer:

  • 1 cm

Step-by-step explanation:

<u>Sides of the picture added margin:</u>

  • l = 8 + 2x
  • w = 5 + 2x
  • A(margin) = 30 cm²

<u>Solution</u>:

  • A(margin) = Total area - A(picture)
  • (8 + 2x)(5 + 2x) - 5*8 = 30
  • 4x² + 10x + 16x + 40 - 40 - 30 = 0
  • 4x² + 26x - 30 = 0
  • 2x² + 13x - 15 = 0
  • x = (-13 + √(13² + 2*4*15))/4
  • x = (-13 + 17)/4
  • x = 1 cm

Note. The other root is ignored as negative.

3 0
3 years ago
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