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lora16 [44]
3 years ago
11

Explain how you can use a quick picture to find 3× 2.7​

Mathematics
2 answers:
Mrac [35]3 years ago
7 0

Answer:

see the picture 8.1 is answer

pogonyaev3 years ago
6 0
Start by drawing 3 squares and using 2.7 as the perimeter.
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Estimate 4,302 + 5,289
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Answer:

9,591

Step-by-step explanation:

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Whats the quake root of 6
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I don’t get what you mean. Reword it
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3 years ago
Amanufacturer of potato chips would like to know whether its bag filling machine works correctly at the 433 gram setting. It is
ankoles [38]

Answer:

There is not enough evidence to support the claim that the bags are under filled.

Step-by-step explanation:

Given :

Population mean, μ = 433

Sample size, n = 26

xbar = 427

Variance, s² = 324 ; Standard deviation, s = √324 = 18

The hypothesis :

H0 : μ = 433

H0 : μ < 433

The test statistic :

(xbar - μ) ÷ (s/√(n))

(427 - 433) / (18 / √26)

-6 / 3.5300904

T = -1.70

The Pvalue :

df = 26-1 = 25 ; α = 0.05

Pvalue = 0.0508

Since Pvalue > α ; WE fail to reject the Null and conclude that there is not enough evidence to support the claim that the bags are underfilled

3 0
3 years ago
A breeder reactor converts uranium-238 into an isotope of plutonium-239 at a rate proportional to the amount of uranium-238 pres
topjm [15]

Let U(t) denote the amount of uranium-238 in the reactor at time t. As conversion to plutonium-239 occurs, the amount of uranium will decrease, so the conversion rate is negative. Because the rate is proportional to the current amount of uranium, we have

\dfrac{\mathrm dU}{\mathrm dt}=-kU

where k>0 is constant. Separating variables and integrating both sides gives

\dfrac{\mathrm dU}U=-k\,\mathrm dt\implies\ln|U|=-kt+C\implies U=Ce^{-kt}

Suppose we start some amount u. This means that at time t=0 we have U(0)=u, so that

u=Ce^{-0k}\implies C=u\implies U=ue^{-kt}

We're given that after 10 years, 99.97% of the original amount of uranium remains. This means (if t is taken to be in years) for some starting amount u,

0.9997u=ue^{-10k}\implies k=-\dfrac{\ln(0.9997)}{10}

The half-life is the time t_{1/2} it takes for the starting amount u to decay to half, 0.5u:

0.5u=ue^{-kt_{1/2}}\implies t_{1/2}=-\dfrac{\ln(0.5)}k=\dfrac{10\ln2}{\ln(0.9997)}

or about 23,101 years. Notice that it doesn't matter what the actual starting amount is, the half-life is independent of that.

7 0
3 years ago
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