A rectangle has an area of 240 square meters and a width of 15 meters.

squares in Step n. f (n) = 8 + 3(n – 1} for n > 1 /(1) = 8, /{n2) = 3+f (n – 1) for n > 2 01)= 8, 7 (n) = 8= ƒ(n=1) forn> 2 Df1)= 3 -8 (n- 1) forn > 1 Of (n) - 37 + 5 for n > 1 32+5 for n>1 CS (n) 3+ an forn 1

Step-by-step explanation:

3 0

1 year ago

Find the area under the standard normal curve to the left of z=1.25.

snow_tiger [21]

For quite some time now, calculators have had statistical functions built in. Here we need to use the stat. function normalcdf, which has only two inputs when we're working with z scores instead of raw scores.

Here,

normalcdf(-100,1.25) = 0.894

This same result could be obtained using a table of z-scores.

3 0

2 years ago

Several years ago, 38% of parents who had children in grades K-12 were satisfied with the quality of education the students r

Colt1911 [192]

Answer:

$0.428 - 1.96\sqrt{\frac{0.428(1-0.428)}{1165}}=0.3995$

$0.428 + 1.96\sqrt{\frac{0.428(1-0.428)}{1165}}=0.4564$

We are confident that the true proportion of people satisfied with the quality of education the students receive is between (0.3995, 0.4564), since the lower value for this confidence level is higher than 0.38 we have enough evidence to conclude that the parents' attitudes toward the quality of education have changed.

Step-by-step explanation:

For this case we are interesting in the parameter of the true proportion of people satisfied with the quality of education the students receive

The confidence level is given 95%, the significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical values are:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

The estimated proportion of people satisfied with the quality of education the students receive is given by:

$\hat p =\frac{499}{1165}= 0.428$

The confidence interval for the proportion if interest is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

And replacing the info given we got:

$0.428 - 1.96\sqrt{\frac{0.428(1-0.428)}{1165}}=0.3995$

$0.428 + 1.96\sqrt{\frac{0.428(1-0.428)}{1165}}=0.4564$

We are confident that the true proportion of people satisfied with the quality of education the students receive is between (0.3995, 0.4564), since the lower value for this confidence level is higher than 0.38 we have enough evidence to conclude that the parents' attitudes toward the quality of education have changed.

8 0

3 years ago