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3 years ago

Remind me to never trust my gut.

Mathematics

2 answers:

netineya [11]3 years ago

8 0

Answer:

1.758 Hope this helps :,D haven't spoken to u in a it tho- want to rp?

Step-by-step explanation:

k0ka [10]3 years ago

3 0

Answer:

Step-by-step explanation:

Distribute the negative sign to turn the equation into pure addition. Then, change the denominator of the two fractions to the same thing.

Multiply 12/11 by 3/3

$\frac{12}{11}*\frac{3}{3}=\frac{36}{33}$

Multiply 2/3 by 11/11

$\frac{2}{3}*\frac{11}{11} =\frac{22}{33}$

Add them together now

$\frac{36}{33} +\frac{22}{33} =\frac{58}{33}$

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A student researcher compares the heights of men and women from the student body of a certain college in order to estimate the d

maria [59]

Answer:

(0.6613, 8.3387)

Step-by-step explanation:

Let the heights of men be the first population and the heights of women be the second population. Then

We have $n_{1} = 15$ , $\bar{x}_{1} = 69.4$ , $s_{1} = 3.09$ and

$n_{2} = 7$ , $\bar{x}_{2} = 64.9$ , $s_{2} = 2.58$ . The pooled

estimate is given by

$s_{p}^{2} = \frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2} = \frac{(15-1)(3.09)^{2}+(7-1)(2.58)^{2}}{15+7-2} = 8.68$

The 99% confidence interval for the true mean difference between the mean height of the men and the mean height of the women is given by (if the samples are independent)

$(\bar{x}_{1}-\bar{x}_{2})\pm t_{0.01/2}s_{p}\sqrt{\frac{1}{15}+\frac{1}{7}}$ , i.e.,

$(69.4-64.9)\pm t_{0.005}2.946\sqrt{\frac{1}{15}+\frac{1}{7}}$

where $t_{0.005}$ is the 0.5th quantile of the t distribution with (15+7-2) = 20 degrees of freedom. So

$4.5\pm(-2.845)(2.946)(0.458)$ , i.e.,

(0.6613, 8.3387)

5 0

3 years ago

Two sevenths of the crowd cheered Wiley. The rest of the crowd stood quietly. What fraction of the crowd stood quietly.

Aleonysh [2.5K]

2/7 - 7/7 = 5/7

5/7 of the crowd stood quietly

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4 years ago

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3 years ago

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skad [1K]

Answer:

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3 years ago

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never [62]

Answer:

107.5 cm

Step-by-step explanation:

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3 years ago