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strojnjashka [21]
3 years ago
14

(suod P.) 240/29 US wa Qures a sus sed pape SeaRJOAM nguo send

Mathematics
1 answer:
romanna [79]3 years ago
4 0

Answer:

English....

Step-by-step explanation:

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5/8 - 1/6 in simplest form
anzhelika [568]
First, you got to answer the question. The answer is 11/24, since 11 is a prime number that's your answer.
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3 years ago
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Assume that y varies inversely with x. If y=1/3 when x=1/2, find y when x=1/4.
LenKa [72]

Answer:

y = \frac{2}{3}

Step-by-step explanation:

Given that y varies inversely with x then the equation relating them is

y = \frac{k}{x} ← k is the constant of variation

To find k use the condition y = \frac{1}{3} when x = \frac{1}{2} , thus

\frac{1}{3} = \frac{k}{\frac{1}{2} } = 2k ( divide both sides by 2 )

k = \frac{1}{6}

y = \frac{1}{6x} ← equation of variation

When x = \frac{1}{4} , then

y = \frac{1}{6(\frac{1}{4}) } = \frac{1}{\frac{3}{2} } = \frac{2}{3}

3 0
3 years ago
Find the greatest common factor.
jonny [76]

Answer:

Greatest is the multiplication exponent by algebra times by the quotient

Step-by-step explanation:

8 0
3 years ago
How do I factor this problem
Keith_Richards [23]

Answer:

−(3x−2)(7x+4)

Step-by-step explanation:

4 0
3 years ago
A random sample of 28 statistics tutorials was selected from the past 5 years and the percentage of students absent from each on
SVEN [57.7K]

Answer:

Step-by-step explanation:

Hello!

X: number of absences per tutorial per student over the past 5 years(percentage)

X≈N(μ;σ²)

You have to construct a 90% to estimate the population mean of the percentage of absences per tutorial of the students over the past 5 years.

The formula for the CI is:

X[bar] ± Z_{1-\alpha /2} * \frac{S}{\sqrt{n} }

⇒ The population standard deviation is unknown and since the distribution is approximate, I'll use the estimation of the standard deviation in place of the population parameter.

Number of Absences 13.9 16.4 12.3 13.2 8.4 4.4 10.3 8.8 4.8 10.9 15.9 9.7 4.5 11.5 5.7 10.8 9.7 8.2 10.3 12.2 10.6 16.2 15.2 1.7 11.7 11.9 10.0 12.4

X[bar]= 10.41

S= 3.71

Z_{1-\alpha /2}= Z_{0.95}= 1.645

[10.41±1.645*(\frac{3.71}{\sqrt{28} } )]

[9.26; 11.56]

Using a confidence level of 90% you'd expect that the interval [9.26; 11.56]% contains the value of the population mean of the percentage of absences per tutorial of the students over the past 5 years.

I hope this helps!

7 0
3 years ago
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