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Ann [662]
2 years ago
7

9.

Mathematics
2 answers:
devlian [24]2 years ago
8 0

Answer:

Step-by-step explanation:

hmm

LenaWriter [7]2 years ago
4 0
The answer is A, y = (x+3)^2 -3
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_____ data are information about people, places, and things that can be measured with numbers.
Westkost [7]

Answer:

numerical

Step-by-step explanation:

i guess not sure though

5 0
3 years ago
ANswer this question!!!!!
4vir4ik [10]

<u><em>Its Going to Be Table B! Good Luck :)⇔</em></u>


6 0
3 years ago
List 3 equivalent ratios of 5/2
jek_recluse [69]
3 equivalent ratios or 5/2
10 /4
20/8
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6 0
2 years ago
Read 2 more answers
This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive an
maxonik [38]

Answer:

5.0 ft-lbf

Step-by-step explanation:

The force is

F = \dfrac{9}{6^x}

This force is not a constant force. For a non-constant force, the work done, <em>W</em>, is

W = \int\limits^{x_2}_{x_1} {F(x)} \, dx

with x_1 and x_2 the initial and final displacements respectively.

From the question, x_1  =0 and x_2 = 12.

Then

W = \int\limits^{12}_0 {\dfrac{9}{6^x}} \, dx

Evaluating the indefinite integral,

\int\limits \dfrac{9}{6^x} \, dx  =9 \int\limits\!\left(\frac{1}{6}\right)^x \, dx

From the rules of integration,

\int\limits a^x\, dx = \dfrac{a^x}{\ln a}

9 \int\limits \left(\frac{1}{6}\right)^x \, dx = 9\times\dfrac{(1/6)^x}{\ln(1/6)} = -5.0229\left(\dfrac{1}{6}\right)^x

Returning the limits,

\left.-5.0229\left(\dfrac{1}{6}\right)^x\right|^{12}_0 = -5.0229(0.1667^{12} - 0.1667^0) = 5.0229 \approx 5.0 \text{ ft-lbf}

4 0
3 years ago
Use the given values of n and p to find the minimum usual value μ−2σ and the maximum usual value μ+2σ. Round to the nearest hund
lbvjy [14]

Answer:

μ−2σ = 1,089.26

μ+2σ = 1,097.62

Step-by-step explanation:

The standard deviation of a sample of size 'n' and proportion 'p' is:

\sigma=\sqrt{\frac{p*(1-p)}{n} }

If n=1139 and p =0.96, the standard deviation is:

\sigma=\sqrt{\frac{p*(1-p)}{n}}\\\sigma = 0.001836

The minimum and maximum usual values are:

\mu-2\sigma = (p-2\sigma)*n\\\mu+2\sigma = (p+2\sigma)*n

\mu-2\sigma = (0.96-2*0.001836)*1139\\\mu-2\sigma = 1,089.26\\\mu+2\sigma = (0.96+2*0.001836)*1139\\\mu+2\sigma = 1,097.62

5 0
2 years ago
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