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liq [111]
3 years ago
6

What is the nth term of the sequence 19, 15.5, 12, 8.5, 5

Mathematics
1 answer:
Nikitich [7]3 years ago
3 0

Answer: 6.5 and 6 is the 9th term

Step-by-step explanation:

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The surface area of a right cone is 400.2 m2. The radius of the cone is 6.0 m. Determine the height of the cone to the nearest m
svetoff [14.1K]
                 S = πr(r + √(h² + r²))
          400.2 = 3.14(6)(6 + √(h² + 6²))
          400.2 = 18.84(6 + √(h² + 36))
          18.84                 18.84
        21¹⁰⁹/₄₇₁ = 6 + √(h² + 36))
        - 6         - 6
        15¹⁰⁹/₄₇₁ = √(h² + 36)
231²²¹⁰⁰⁵/₂₂₁₈₄₁ = h² + 36
- 36                       - 36
195²²¹⁰⁰⁵/₂₂₁₈₄₁ = h²
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8 0
3 years ago
The solutions to the inequality y ≤ 2x − 4 are shaded on the graph. Which point is a solution?
I am Lyosha [343]

Answer:

it's the point (3, 2)

Step-by-step explanation:

Plug in 3 into x and plug in 2 into y. You will get 2≤2(3)-4. This ends up equalling 2≤2. THis is the answer cuz 2 is equal to or less than 2. :)

8 0
3 years ago
Which of the diagrams below represents the statement if it is a rectangle, then it is a square
Naily [24]
All the rectangle are square if length becomes equal to breath !
4 0
3 years ago
What is 10x10, please help xd
Juliette [100K]

Answer:

100 lol

Step-by-step explanation:

ty for the points! <3

7 0
3 years ago
Read 2 more answers
To obtain information on the corrosion-resistance properties of a certain type of steel conduit, 45 specimens are buried in soil
Fiesta28 [93]

Answer:

The sample has not met the required specification.

Step-by-step explanation:

As the average of the sample suggests that the true average penetration of the sample could be greater than the 50 mils established, we formulate our hypothesis as follow

H_0: The true average penetration is 50 mils

H_a: The true average penetration is > 50 mils

Since we are trying to see if the true average is greater than 50, this is a right-tailed test.

If the <em>level of confidence</em> is α = 0.05 then the z_\alpha score against we are comparing with, is 1.64 (this is because the area under the normal curve N(0;1) to the right of 1.64 is 0.05)

The z-score associated with this test is

z=\frac{\bar x-\mu}{s/\sqrt{n}}

where

\bar x = <em>mean of the sample</em>

\mu = <em>average established by the specification</em>

s = <em>standard deviation of the sample</em>

n = <em>size of the sample</em>

Computing this value of z we get z = 3.42

Since z >z_\alpha we can conclude that the sample has not met the required specification.

5 0
3 years ago
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