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QveST [7]
2 years ago
10

Second piece help ASAPP..

Mathematics
2 answers:
Evgesh-ka [11]2 years ago
7 0

Answer:

#2 80

Step-by-step explanation:

you start from the back and you have to put the place value

Ivan2 years ago
6 0

Answer:

I can help you but I'm not see it clear

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Y=-4x-2 and intersects at the points (4,-1)
rosijanka [135]

Answer:

It doesn't intersect at that point

{ \bf{y =  - 4x - 2}} \\ y =  - (4 \times 4) - 2 \\ y =  - 18

4 0
3 years ago
Read 2 more answers
Combien font 2 + 2 - 5
wel

Hello!

2 + 2 - 5 = -1.

I hope it helps!

3 0
3 years ago
Find the prime factorization of each number.<br> 1. 50 =
Varvara68 [4.7K]

Answer:

1.5 is not a prime number

Step-by-step explanation:

7 0
3 years ago
Evaluate the given integral by changing to polar coordinates. 8xy dA D , where D is the disk with center the origin and radius 9
BabaBlast [244]

Answer:

0

Step-by-step explanation:

∫∫8xydA

converting to polar coordinates, x = rcosθ and y = rsinθ and dA = rdrdθ.

So,

∫∫8xydA = ∫∫8(rcosθ)(rsinθ)rdrdθ = ∫∫8r²(cosθsinθ)rdrdθ = ∫∫8r³(cosθsinθ)drdθ

So we integrate r from 0 to 9 and θ from 0 to 2π.

∫∫8r³(cosθsinθ)drdθ = 8∫[∫r³dr](cosθsinθ)dθ

= 8∫[r⁴/4]₀⁹(cosθsinθ)dθ

= 8∫[9⁴/4 - 0⁴/4](cosθsinθ)dθ

= 8[6561/4]∫(cosθsinθ)dθ

= 13122∫(cosθsinθ)dθ

Since sin2θ = 2sinθcosθ, sinθcosθ = (sin2θ)/2

Substituting this we have

13122∫(cosθsinθ)dθ = 13122∫(1/2)(sin2θ)dθ

= 13122/2[-cos2θ]/2 from 0 to 2π

13122/2[-cos2θ]/2 = 13122/4[-cos2(2π) - cos2(0)]

= -13122/4[cos4π - cos(0)]

= -13122/4[1 - 1]

= -13122/4 × 0

= 0

5 0
3 years ago
Lucy is using a one-sample t ‑test based on a simple random sample of size n = 22 to test the null hypothesis H 0 : μ = 16.000 c
slega [8]

Answer:

The value of test statistic is 1.338

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = 16.000

Sample mean, \bar{x} = 16.218

Sample size, n = 22

Alpha, α = 0.05

Sample standard deviation, s = 0.764

First, we design the null and the alternate hypothesis

H_{0}: \mu = 16.000\text{ cm}\\H_A: \mu < 16.000\text{ cm}

We use one-tailed t test to perform this hypothesis.

Formula:

t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }

Putting all the values, we have

t_{stat} = \displaystyle\frac{16.218 - 16.000}{\frac{0.764}{\sqrt{22}} } = 1.338

Thus, the value of test statistic is 1.338

4 0
3 years ago
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