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vfiekz [6]
3 years ago
8

Need help plzzzzzzzz

Mathematics
1 answer:
Igoryamba3 years ago
8 0
Box 1: A and J

Box 2: Point B

Box 3: Sorry I cant see it. Can you please send it again?

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(–2, 18), (0, 10), (2, 2)
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What is the value of x if Y=log x & If y=10?
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x = 10^{10}

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Under which conditions do you expect helium gas to deviate most from ideal behavior?
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"300 K and 2 atm" is the condition among the conditions given in the question that you can <span>expect helium gas to deviate most from ideal behavior. The correct option among all the options that are given in the question is the second option or option "b". I hope the answer has helped you.</span>
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Calculate the sample mean and sample variance for the following frequency distribution of heart rates for a sample of American a
spin [16.1K]

Answer:

Mean = 68.9

s^2 =18.1 --- Variance

Step-by-step explanation:

Given

\begin{array}{cccccc}{Class} & {51-58} & {59-66} & {67-74} & {75-82} & {83-90} \ \\ {Frequency} & {6} & {3} & {11} & {13} & {4} \ \end{array}

Solving (a): Calculate the mean.

The given data is a grouped data. So, first we calculate the class midpoint (x)

For 51 - 58.

x = \frac{1}{2}(51+58) = \frac{1}{2}(109) = 54.5

For 59 - 66

x = \frac{1}{2}(59+66) = \frac{1}{2}(125) = 62.5

For 67 - 74

x = \frac{1}{2}(67+74) = \frac{1}{2}(141) = 70.5

For 75 - 82

x = \frac{1}{2}(75+82) = \frac{1}{2}(157) = 78.5

For 83 - 90

x = \frac{1}{2}(83+90) = \frac{1}{2}(173) = 86.5

So, the table becomes:

\begin{array}{cccccc}{x} & {54.5} & {62.5} & {70.5} & {78.5} & {86.5} \ \\ {Frequency} & {6} & {3} & {11} & {13} & {4} \ \end{array}

The mean is then calculated as:

Mean = \frac{\sum fx}{\sum f}

Mean = \frac{54.5*4+62.5*3+70.5*11+78.5*13+86.5*4}{6+3+11+13+4}

Mean = \frac{2547.5}{37}

Mean = 68.9 -- approximated

Solving (b): The sample variance:

This is calculated as:

s^2 =\frac{\sum (x - \overline x)^2}{\sum f - 1}

So, we have:

s^2 =\frac{(54.5-68.9)^2+(62.5-68.9)^2+(70.5-68.9)^2+(78.5-68.9)^2+(86.5-68.9)^2}{37 - 1}

s^2 =\frac{652.8}{36}

s^2 =18.1 -- approximated

5 0
2 years ago
Help please i don’t know how to do it
Norma-Jean [14]

Answer:

Step-by-step explanation:

\frac{dy}{dx}=x^{2}(y-1)\\\frac{1}{y-1} \text{ } dy=x^{2} \text{ } dx\\\int \frac{1}{y-1} \text{ } dy=\int x^{2} \text{ } dx\\\ln|y-1|=\frac{x^{3}}{3}+C\\

From the initial condition,

\ln|3-1|=\frac{0^{3}}{3}+C\\\ln 2=C

So we have that \ln |y-1|=\frac{x^{3}}{3}+\ln 2\\e^{\frac{x^{3}}{3}+\ln 2}=y-1\\2e^{\frac{x^{3}}{3}}=y-1\\y=2e^{\frac{x^{3}}{3}}+1

4 0
2 years ago
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