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kicyunya [14]
3 years ago
13

The 10th term of an arithmetic proression is 68 and the common difference is 7 find the first term of the sequence

Mathematics
1 answer:
leonid [27]3 years ago
3 0

Answer:

5

Step-by-step explanation:

first term + (10-1)7=68

first term= 68- 63=5

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Jr tripled the area of his garden by increasing the length by 38. What are the original dimensions
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I confused ! I don't get what your trying to say please explain better !
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For a continuous random variable X, P(20 ≤ X ≤ 40) = 0.15 and P(X > 40) = 0.16. Calculate the following probabilities. (Leave
yan [13]

Answer:

For a continuous random variable X, P(20 ≤ X ≤ 40) = 0.15 and P(X > 40) = 0.16.

Step-by-step explanation:

Here, P(x > 40) = 0.16

a). P(x < 40) = 1 - P(x > 40)

                   =  1 - 0.16

                    = 0.84

b). P(x < 20) = 1 - P(x\geq 20)

                    = 1 - {P(20 ≤ X ≤ 40) + P(X > 40)}

                    = 1 - (0.15 + 0.16 )  

                    = 1 - 0.31

                    = 0. 69

c). P(x = 40) = 0; The probability that a continuous variable assume a particular value is zero.

5 0
3 years ago
URGENT HELP!!!!<br> Picture included
natka813 [3]

Answer:

Length (L) = 72 feet

Step-by-step explanation:

From the question given above, the following data were obtained:

Period (T) = 9.42 s

Pi (π) = 3.14

Length (L) =?

The length of the pendulum can be obtained as follow:

T = 2π √(L/32)

9.42 = (2 × 3.14) √(L/32)

9.42 = 6.28 √(L/32)

Divide both side by 6.28

√(L/32) = 9.42 / 6.28

Take the square of both side

L/32 = (9.42 / 6.28)²

Cross multiply

L = 32 × (9.42 / 6.28)²

L = 72 feet

Thus, the Lenght is 72 feet

5 0
3 years ago
PLEASE HURRY IM TIMED!!!
andriy [413]

<u>Answer-</u>

<em>A. Brandon’s sound intensity level is 1/4th as compared to Ahmad’s.</em>

<u>Solution-</u>

Given that, loudness measured in dB is

L=10\log \frac{I}{I_0}

Where,

I   = Sound intensity,

I₀ = 10⁻¹² and is the least intense sound a human ear can hear

Given in the question,

I₁ = Intensity at Brandon's = 10⁻¹⁰

I₂ = Intensity at Ahmad's  = 10⁻⁴

Then,

L_1=10\log \frac{I_1}{I_0}=10\log \frac{10^{-10}}{10^{-12}}=10\log \frac{1}{10^{-2}}=10\log 10^{2}=2\times10\log 10=20

L_2=10\log \frac{I_2}{I_0}=10\log \frac{10^{-4}}{10^{-12}}=10\log \frac{1}{10^{-8}}=10\log 10^{8}=8\times10\log 10=80

\therefore \frac{L_1}{L_2} =\frac{20}{80} =\frac{1}{4}

5 0
3 years ago
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