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2 years ago

Evaluate-2 x 3/12 x 5/10 x 7/15

1 answer:

7 0

$\displaystyle\Large \boldsymbol -2 \cdot \frac{3}{12} \cdot \frac{5}{10} \cdot \frac{7}{15} =-2 \cdot \frac{3 \!\!\!\!\diagup \cdot1}{3 \!\!\!\!\diagup\cdot 4} \cdot \frac{5 \!\!\!\!\diagup\cdot 1}{5 \!\!\!\!\diagup\cdot 2} \cdot \frac{7}{15} =\\\\\\-\frac{2 \!\!\!\!\diagup\cdot 7 }{2 \!\!\!\!\diagup\cdot 4 \cdot 15} =\boxed{-\frac{7}{60}}$

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Consider two functions: g(x)=−x2−6x and the quadratic function f(x) shown in the table.

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Hence average rate of change is greater for f(x) in (0,3).

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To find average rate of change of f and g in (0,3)

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Step-by-step explanation:

The hypothesis for this test can be defined as follows:

H₀: Men do not spend more than women on St. Patrick's day, i.e. μ₁ = μ₂.

Hₐ: Men spend more than women on St. Patrick's day, i.e. μ₁ > μ₂.

The population standard deviations are not known.

So a t-distribution will be used to perform the test.

The test statistic for the test of difference between mean is:

$t=\frac{\bar x_{1}-\bar x_{2}}{\sqrt{\frac{s^{2}_{1}}{n_{1}}+\frac{s^{2}_{2}}{n_{2}}} }$

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$\bar x_{1}=55\\s_{1}=18\\n_{1}=90\\\bar x_{1}=44\\s_{1}=16\\n_{1}=86$

Compute the value of the test statistic as follows:

$t=\frac{\bar x_{1}-\bar x_{2}}{\sqrt{\frac{s^{2}_{1}}{n_{1}}+\frac{s^{2}_{2}}{n_{2}}} }=\frac{55-44}{\sqrt{\frac{15^{2}}{90}+\frac{16^{2}}{86} }}=4.70$

Thus, the value of the test statistic is 4.70.

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3 years ago

Evaluate-2 x 3/12 x 5/10 x 7/15​

Evaluate-2 x 3/12 x 5/10 x 7/15