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muminat
3 years ago
9

Which number is the best approximation for 52√+23√ ?

Mathematics
1 answer:
Agata [3.3K]3 years ago
6 0
<span>52√+23√ = 4.79

Approx 5.1 is the answer.

Hope that helps. -UF aka Nadia
</span>
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5.<br> Which picture shows a rotation of the flag?<br> (1)<br> (2)<br> (3)<br> (4)
RUDIKE [14]

Answer:

4 my good sir

Step-by-step explanation:

4 0
3 years ago
Best known for its testing program, ACT, Inc., also compiles data on a variety of issues in education. In 2004 the company repor
GarryVolchara [31]

Answer:

\mu_p -\sigma_p = 0.74-0.0219=0.718

\mu_p +\sigma_p = 0.74+0.0219=0.762

68% of the rates are expected to be betwen 0.718 and 0.762

\mu_p -2*\sigma_p = 0.74-2*0.0219=0.696

\mu_p +2*\sigma_p = 0.74+2*0.0219=0.784

95% of the rates are expected to be betwen 0.696 and 0.784

\mu_p -3*\sigma_p = 0.74-3*0.0219=0.674

\mu_p +3*\sigma_p = 0.74+3*0.0219=0.806

99.7% of the rates are expected to be betwen 0.674 and 0.806

Step-by-step explanation:

Check for conditions

For this case in order to use the normal distribution for this case or the 68-95-99.7% rule we need to satisfy 3 conditions:

a) Independence : we assume that the random sample of 400 students each student is independent from the other.

b) 10% condition: We assume that the sample size on this case 400 is less than 10% of the real population size.

c) np= 400*0.74= 296>10

n(1-p) = 400*(1-0.74)=104>10

So then we have all the conditions satisfied.

Solution to the problem

For this case we know that the distribution for the population proportion is given by:

p \sim N(p, \sqrt{\frac{p(1-p)}{n}})

So then:

\mu_p = 0.74

\sigma_p =\sqrt{\frac{0.74(1-0.74)}{400}}=0.0219

The empirical rule, also referred to as the three-sigma rule or 68-95-99.7 rule, is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ). Broken down, the empirical rule shows that 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ).

\mu_p -\sigma_p = 0.74-0.0219=0.718

\mu_p +\sigma_p = 0.74+0.0219=0.762

68% of the rates are expected to be betwen 0.718 and 0.762

\mu_p -2*\sigma_p = 0.74-2*0.0219=0.696

\mu_p +2*\sigma_p = 0.74+2*0.0219=0.784

95% of the rates are expected to be betwen 0.696 and 0.784

\mu_p -3*\sigma_p = 0.74-3*0.0219=0.674

\mu_p +3*\sigma_p = 0.74+3*0.0219=0.806

99.7% of the rates are expected to be betwen 0.674 and 0.806

5 0
3 years ago
Find two numbers with difference 62 and whose product is a minimum.
nika2105 [10]

Answer:

31 and -31

Step-by-step explanation:

6 0
3 years ago
Water is leaking out of a large barrel at a rate proportional to the square rooot of the depth of the water at that time. If the
sdas [7]

Answer:

It will take about 35.49 hours for the water to leak out of the barrel.

Step-by-step explanation:

Let y(t) be the depth of water in the barrel at time t,  where y is measured in inches and t in hours.

We know that water is leaking out of a large barrel at a rate proportional to the square root of the depth of the water at that time. We then have that

                                                 \frac{dy}{dt}=-k\sqrt{y}

where k is a constant of proportionality.

Separation of variables is a common method for solving differential equations. To solve the above differential equation you must:

Multiply by \frac{1}{\sqrt{y}}

\frac{1}{\sqrt{y}}\frac{dy}{dt}=-k

Multiply by dt

\frac{1}{\sqrt{y}}\cdot dy=-k\cdot dt

Take integral

\int \frac{1}{\sqrt{y}}\cdot dy=\int-k\cdot dt

Integrate

2\sqrt{y}=-kt+C

Isolate y

y(t)=(\frac{C}{2} -\frac{k}{2}t)^2

We know that the water level starts at 36, this means y(0)=36. We use this information to find the value of C.

36=(\frac{C}{2} -\frac{k}{2}(0))^2\\C=12

y(t)=(\frac{12}{2} -\frac{k}{2}t)^2\\\\y(t)=(6 -\frac{k}{2}t)^2

At t = 1, y = 34

34=(6 -\frac{k}{2}(1))^2\\k=12-2\sqrt{34}

So our formula for the depth of water in the barrel is

y(t)=(6 -\frac{12-2\sqrt{34}}{2}t)^2\\\\y(t)=\left(6-\left(6-\sqrt{34}\right)t\right)^2\\

To find the time, t, at which all the water leaks out of the barrel, we solve the equation

\left(6-\left(6-\sqrt{34}\right)t\right)^2=0\\\\t=3\left(6+\sqrt{34}\right)\approx 35.49

Thus, it will take about 35.49 hours for the water to leak out of the barrel.

5 0
4 years ago
Whats the answer to 10×(2.5+13.5)
oee [108]

Simplify 2.5 + 13.5 to 16

10 × 16

Simplify

160

7 0
3 years ago
Read 2 more answers
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