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blagie [28]
2 years ago
13

Which are vertical angles? O AFE and BFD O BFC and DFE O AFE and CFD O BFC and EFA

Mathematics
1 answer:
mel-nik [20]2 years ago
3 0

the correct answer is a, because a vertical angle is made of 2 intercepting lines.

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Cost of an airplane flight increase from $250 to $260. What is the percentage increase?​
Sedaia [141]

Answer:

Solution given:

Increase percentage=\frac{260-250}{250}*100%

=4%

<u>the percentage </u><u>increase</u><u> </u><u>is</u><u> </u><u>4</u><u>%</u><u>.</u>

3 0
2 years ago
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Vadim26 [7]
1 gallon = 16 cups. so 16x3.5=56. so 56 cups
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A research company claims that no more than 55% of Americans regularly watch FOX News. You decide to test this claim and ask a r
7nadin3 [17]

Answer: z(e) = 2.07

Step-by-step explanation:

1.-The problem is about a test of proportions. As research company claims than no more 55 % of Americans regularly watch Fox News

The null hypothesis is  H₀       P₀ ≤ 0.55       from 55%

And the alternative hypothesis is  Hₐ      Pₐ > .55

Is one tail test

2.-We have to specify significance level we assume our test will be for a significance level α = 5%  or α = 0,05

3.-Calculation of z (c) = ??  and z (e) = '??

For z (c) we find in z table the value of  z(c) = 1.64

For z (e) = ( P -P₀)/√p₀q₀/n     z(e) = 0.05 / √(0.55*0,45)/425                         z(e) = 0,05/ 0.02413    z(e) = 2.07

z(e) > z(c)  threfore z(e) is in the rejection zone . We reject null hyothesis

4 0
3 years ago
Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating
natulia [17]

Answer:

V=25088π vu

Step-by-step explanation:

Because the curves are a function of "y" it is decided to take the axis of rotation as y

, according to the graph 1 the cutoff points of f(y)₁ and f(y)₂ are ±2

f(y)₁ = 7y²-28;  f(y)₂=28-7y²

y=0;   x=28-0 ⇒ x=28

x=0;   0 = 7y²-28 ⇒ 7y²=28 ⇒ y²= 28/7 =4 ⇒ y=√4 =±2

Knowing that the volume of a solid of revolution  V=πR²h, where R²=(r₁-r₂) and h=dy then:

dV=π(7y²-28-(28-7y²))²dy ⇒dV=π(7y²-28-28+7y²)²dy = 4π(7y²-28)²dy

dV=4π(49y⁴-392y²+784)dy integrating on both sides

∫dV=4π∫(49y⁴-392y²+784)dy ⇒ solving  ∫(49y⁴-392y²+784)dy

49∫y⁴dy-392∫y²dy+784∫dy =

V=4π( 49\frac{y^{5} }{5}-392\frac{y^{3}}{3}+784y ) evaluated -2≤y≤2, or 2(0≤y≤2), also

V=8\pi(49\frac{2^{5} }{5}-392\frac{2^{3} }{3}+784.2)  ⇒ V=25088π vu

8 0
3 years ago
Solution set for 6 tan 3x=6
choli [55]
6\tan3x=6\ \ \ \ |divide\ both\ sides\ by\ 6\\\\\tan3x=1`\\\\3x=\dfrac{\pi}{4}+k\pi\ \ \ \ \ |divide\ both\ sides\ by\ 3\\\\\boxed{x=\dfrac{\pi}{12}+\dfrac{k\pi}{3}\ where\ k\in\mathbb{Z}}
8 0
3 years ago
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