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2 years ago

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

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3 years ago

ANswer this question!!!!!

4vir4ik [10]

Its Going to Be Table B! Good Luck :)⇔

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3 years ago

List 3 equivalent ratios of 5/2

jek_recluse [69]

3 equivalent ratios or 5/2
10 /4
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30/12

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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, W, 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