Find the circumference of the object. Use 3.14 or 22 for . Round to the nearest hundredth if necessary

At a company picnic, 1/2 of the people are employees. 2/5 are employees' spouses. What percent are neither?

kupik [55]

Answer:

10% of the people at the picnic are neither.

Step-by-step explanation:

To get the answer,

1/2 + 2/5 = 9/10

10/10 - 9/10 = 1/10

1/10 = 10%

Thanks, I hope I got this right!

7 0

3 years ago

Consider the accompanying data on breaking load (kg/25mm width) for various fabrics in both an unabraded condition and an abrade

Hitman42 [59]

Answer:

a) $t=\frac{(43.625-36.375)-0}{\sqrt{\frac{9.694^2}{8}+\frac{11.253^2}{8}}}}=1.38$

$p_v = P(t_{14}>1.38) =0.0946$

Since the p value for this case is higher than the significance level of 0.01 we have enough evidence to FAIL to reject the null hypothesis and we can conclude that we don't have significant differences between the true means

b) $(43.625 -36.375) - 2.98 \sqrt{\frac{9.694^2}{8} +\frac{11.253^2}{8}} = -23.976$

$(43.625 -36.375) + 2.98 \sqrt{\frac{9.694^2}{8} +\frac{11.253^2}{8}} = 38.476$

Step-by-step explanation:

We have the following data given:

U 36.4 55.0 51.5 38.7 43.2 48.8 25.6 49.8

A 28.5 20.0 46.0 34.5 36.5 52.5 26.5 46.5

$\bar X_{U}=43.625$ represent the mean for sample U

$\bar X_{A}=36.375$ represent the mean for sample A

$s_{U}=9.694$ represent the sample standard deviation for U

$s_{A}=11.253$ represent the sample standard deviation for A

$n_{U}=8$ sample size for the group U

$n_{A}=8$ sample size for the group A

$\alpha=0.01$ Significance level provided

t would represent the statistic

Part a

For this case we want to try test if the average breaking load of various fabrics in unabraded condition is greater than that in abraded, so then the hypothesis are:

Null hypothesis: $\mu_{U}-\mu_{A} \leq 0$

Alternative hypothesis: $\mu_{U} - \mu_{A}> 0$

The statistic for this case is given by:

$t=\frac{(\bar X_{U}-\bar X_{A})-\Delta}{\sqrt{\frac{s^2_{U}}{n_{U}}+\frac{s^2_{A}}{n_{A}}}}$ (1)

And the degrees of freedom are given by :

$df = n_U + n_A -2= 8+8-2 =14$

The statistic for this case would be:

$t=\frac{(43.625-36.375)-0}{\sqrt{\frac{9.694^2}{8}+\frac{11.253^2}{8}}}}=1.38$

P value

The p value can be calculated like this:

$p_v = P(t_{14}>1.38) =0.0946$

Since the p value for this case is higher than the significance level of 0.01 we have enough evidence to FAIL to reject the null hypothesis and we can conclude that we don't have significant differences between the true means

Part b

The confidence interval for the true average difference would be given by:

$(\bar X_U -\bar X_A) \pm t_{\alpha/2} \sqrt{\frac{s^2_U}{n_U} +\frac{s^2_A}{n_A}}$