I really need help,please. also happy holidays!

Please help !!! I’m stuck

Kruka [31]

Answer:

2/3 ≈ 0.6667

Step-by-step explanation:

apply soh cah toa

sine=opposite/hypo

sine=4/6

sine=2/3

4 0

3 years ago

In the straightedge and compass construction of the parallel line below, which of the following reasons can you use to prove tha

Eva8 [605]

Answer:

D.

Step-by-step explanation:

Lines EG and CD are cut by transversal CF.

By construction, ∠FEG=∠FCD. These two angles are corresponding angles.

Since two corresponding angles are congruent, then lines EG and CD are parallel (by converse of the corresponding angles postulate).

Converse of the Corresponding Angles Postulate: If the corresponding angles formed by two lines and a transversal are congruent, then lines are parallel.

3 0

3 years ago

Sea un cuadrado de 2 pulgadas de lado uniendo los puntos medios se obtiene otro cuadrado inscrito en el anterior si repetimos es

Ne4ueva [31]

Answer:

1) La serie geométrica formada es

4, 2, 1,..., ∞

2) La suma al infinito de las áreas de los cuadrados es 8 in.²

Step-by-step explanation:

1) El área del primer cuadrado, a₁ = 2² = 4 pulgadas²

El área del siguiente cuadrado, a₂ = (√ (1² + 1²)) ² = (√2) ² = 2 pulg²

El área del siguiente cuadrado, a₃ = ((√ (2) / 2) ² + (√ (2) / 2) ²) = 1 pulg²

Por lo tanto, la razón común, r = a₂ / a₁ = 2/4 = a₃ / a₂ = 1/2

Las áreas de los cuadrados progresivos forman una progresión geométrica como sigue;

4, 4×(1/2), 4 ×(1/2)²,...,4× $(1/2)^{\infty}$

De donde obtenemos la serie geométrica formada de la siguiente manera;

4, 2, 1,..., ∞

2) La suma de 'n' términos de una progresión geométrica hasta el infinito para -1 <r <1 se da como sigue;

$S_{\infty} = \dfrac{a}{1 - r}$

Por lo tanto, la suma de las áreas de los cuadrados hasta el infinito se obtiene sustituyendo los valores de 'a' y 'r' en la ecuación anterior de la siguiente manera;

$La \ suma \ al \ infinito \ del \ cuadrado \ S_{\infty} = \dfrac{4 \ in.^2}{1 - \dfrac{1}{2} } = \dfrac{4 \ in.^2}{\left(\dfrac{1}{2} \right)} = 2 \times 4 \ in.^2= 8 \ in.^2$

La suma al infinito de las áreas de los cuadrados, $S_{\infty}$ = 8 in.²

7 0

3 years ago

Wyatt is training for a race. He will run 3

Nady [450]

You take 3/4 and multiply that by eight and you’ll get 6

8 0

4 years ago

Read 2 more answers

onsider the following hypothesis test: H 0: 50 H a: > 50 A sample of 50 is used and the population standard deviation is 6. U

kondaur [170]

Answer:

a) z(e) > z(c) 2.94 > 1.64 we are in the rejection zone for H₀ we can conclude sample mean is great than 50. We don´t know how big is the population .We can not conclude population mean is greater than 50

b) z(e) < z(c) 1.18 < 1.64 we are in the acceptance region for H₀ we can conclude H₀ should be true. we can conclude population mean is 50

c) 2.12 > 1.64 and we can conclude the same as in case a

Step-by-step explanation:

The problem is concerning test hypothesis on one tail (the right one)

The critical point z(c) ; α = 0.05 fom z table w get z(c) = 1.64 we need to compare values (between z(c) and z(e) )

The test hypothesis is:

a) H₀ ⇒ μ₀ = 50 a) Hₐ μ > 50 ; for value 52.5

b) Hₐ μ > 50 ; for value 51

c) Hₐ μ > 50 ; for value 51.8

With value 52.5

The test statistic z(e) ??

a) z(e) = ( μ - μ₀ ) /( σ/√50) z(e) = (2.5*√50 )/6 z(e) = 2.94

2.94 > 1.64 we are in the rejected zone for H₀ we can conclude sample mean is great than 50. We don´t know how big is the population .We can not conclude population mean is greater than 50

b) With value 51

z(e) = ( μ - μ₀ ) /( σ/√50) ⇒ z(e) = √50/6 ⇒ z(e) = 1.18

z(e) < z(c) we are in the acceptance region for H₀ we can conclude H₀ should be true. we can conclude population mean is 50

c) the value 51.8

z(e) = ( μ - μ₀ ) /( σ/√50) ⇒ z(e) = (1.8*√50)/ 6 ⇒ z(e) = 2.12

2.12 > 1.64 and we can conclude the same as in case a

8 0

3 years ago