75% of blank peaches is 15 peaches

garik1379 [7]

You would do 2 times 6 and get 12. Then you subtract 12 from -20 and you get -8. so -8 is your answer.

6 0

3 years ago

The height of a flare fired from a 32-foot high platform can be modeled by the function h= -16t^2 + 16t + 32, where h is the hei

Amanda [17]

Answer:

Time, t = 2 seconds

Step-by-step explanation:

The height of a flare fired from a 32-foot high platform can be modeled by the function :

$h= -16t^2 + 16t + 32$

Here, h is the height in feet above the ground and t is the time in seconds.

It is required to find the time taken by the flare to reach the ground. When its reaches ground, height covered by it equals 0 such that,

h = 0

$-16t^2 + 16t + 32=0\\\\t=\dfrac{-16+\sqrt{16^2-4\times (-16)(32)} }{2(-16)}, \dfrac{-16-\sqrt{16^2-4\times (-16)(32)} }{2(-16)}\\\\t=-1\ s, 2\ s$

Neglecting t = -1 seconds, the time taken by the flare to reach the ground is 2 seconds.

4 0

2 years ago

A random sample of 30 households was selected as part of a study on electricity usage, and the number of kilowatt-hours (kWh) wa

grin007 [14]

Answer:

The 98% confidence interval for the mean usage in the March quarter of 2006, in kWh, was (333.87, 416.13).

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 30 - 1 = 29

98% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 29 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.98}{2} = 0.99$ . So we have T = 2.462

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 2.462\frac{91.5}{\sqrt{30}} = 41.13$

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 375 - 41.13 = 333.87 kWh

The upper end of the interval is the sample mean added to M. So it is 375 + 41.13 = 416.13 kWh

The 98% confidence interval for the mean usage in the March quarter of 2006, in kWh, was (333.87, 416.13).

6 0

3 years ago

The distribution of scores on a recent test closely followed a Normal Distribution with a mean of 22 points and a standard devia

maksim [4K]

6.68% of the students scored at least 19 points on this test while the 85 percentile has a test score of 24.08 points

The z score is used to determine by how many standard deviations the raw score is above or below the mean. The z score is given by:

$z=\frac{x-\mu}{\sigma} \\\\where\ \mu=mean.\sigma=standard\ deviation, x=raw\ score\\\\Given\ that\ mean(\mu)=22,standard \ deviation (\sigma)=2\\\\For\ x>19:\\\\z=\frac{22-19}{2}=1.5$