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Eddi Din [679]
3 years ago
13

PLEASE HELP!!! I DONT HAVE ENOUGH POINTS TO POST MORE THAN ONCE!

Mathematics
2 answers:
almond37 [142]3 years ago
8 0

Answer:

-3

Step-by-step explanation:

So, heres hopefully what you're looking for. I answered the last one for you, but i cant promise everything on this one lol. Looking at the table, you decrease on the Y from five to one. Thats moving four times. Then the next part, negative one to one. Thats 2 times, coming for a total of 6 movements. Now, to find average rate of change. 6/2. Thats positive 3, and something is wrong with it. Its missing its negative! Heres why, you are decreasing not only on Y, but on X. During decrease, it NEEDS that negative or the answer is completely incorrect. PLEASE do NOT forget... Rate Of Change does not stay consistant. Every single time your moving from something different than these numbers, you gotta be careful and recalculate it or its completely wrong.

I'm here to help. Give me a shout. Sorry if that wasnt too clear, I tried

Marat540 [252]3 years ago
3 0

Answer:

-3.

Step-by-step explanation:

In the table, the y value goes down by -4 and then -2, for a total of -6.

We did this over a period of two units (To go from -1 to 1, we add 2).

-6/2 = -3.

-3 is the average rate of change over the interval.

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What is the approximate volume of one refraction cup? What’s its formula ?
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Answer:

Volume = 0.1696 L

The formula is Area of base of refraction cup × Depth of the refraction cup

Step-by-step explanation:

The dimensions of a refraction cup are;

Diameter = 120 mm

Depth = 30 mm

Therefore, volume of the refraction cup = Area of base × Depth

Volume of the refraction cup = (π×120²/4)/2 × 30 = 169646.003 mm³

1 L = 1000000 mm³

Therefore, 169646.003 mm³ = 169646.003/1000000 m³ = 0.1696 L

The formula is Area of base of refraction cup × Depth of the refraction cup.

6 0
3 years ago
A civil engineer is analyzing the compressive strength of concrete. Compressive strength is normally distributed with A random s
wel

Answer:

95%: (3278.354 ; 3270.083)

99% : (3221.646 ; 3278.354)

Step-by-step explanation:

Given :

Sample size, n = 12

Mean, xbar = 3250

Sample standard deviation = √1000

The 95% confidence interval :

Mean ± Margin of error

Margin of Error = Tcritical * s/√n

Tcritical at 0.05, df=12-1 = 11 ;

Tcritical at 95% = 2.20

Hence,

Margin of Error = (2.20 * √1000/√12) = 20.083

Confidence interval : 3250 ± 20.083

Lower boundary = 3250 - 20.083 = 3229.917

Upper boundary = 3250 + 20.083 = 3270.083

2.)

The 99% confidence interval :

Mean ± Margin of error

Margin of Error = Tcritical * s/√n

Tcritical at 0.01, df=12-1 = 11 ;

Tcritical at 99% = 3.106

Hence,

Margin of Error = (3.106 * √1000/√12) = 28.354

Confidence interval : 3250 ± 28.354

Lower boundary = 3250 - 28.354 = 3221.646

Upper boundary = 3250 + 28.354 = 3278.354

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Answer:

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Step-by-step explanation:

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