The area of triangle AEB is 24.5 sq units. How long is `segment\ AE`?

“a number n equals 7 more than half the number” can be written as what

sergiy2304 [10]

Answer:

Step-by-step explanation:

n=n/2 +7

3 0

3 years ago

A legislative committee is weighing the pros and cons of a clean-air program. If it spends $1 million on the program, the air wi

Ksivusya [100]

B..
After each million spent, it decreased by 50%.. after 1 million +20% after 2 million +10% .. and so on

3 million is a lot for only 5 percent.. imo

4 0

4 years ago

Read 2 more answers

Simplify: (2x²y²) (-3xy³)<br>please no links and explain please. I need it fast

Shtirlitz [24]

Step-by-step explanation:

Solution:

(6xy) × (-3x²y³)

= {6 × (-3)} × {xy × x²y³}

= -18x1+2 y1+3

= -18x³y⁴.

3 0

3 years ago

A new sales training program has been instituted at a rent-to-own company. Prior to the training, 10 employees were tested on th

Lunna [17]

Answer:

C. H0: µD = 0, HA: µD <0

Step-by-step explanation:

We assume that the paired sample data are simple random samples and that the differences have a distribution that is approximately normal. So for this case is better apply a paired t test.

A paired t-test is used to compare two population means where you have two samples in which observations in one sample can be paired with observations in the other sample. For example if we have Before-and-after observations (This problem) we can use it.

Let put some notation

x=test value for pretest , y = test value for posttest

x: 66, 94, 87, 84, 76, 88

y: 75, 100, 93, 85, 75, 90

The system of hypothesis for this case are:

Null hypothesis: $\mu_D \geq 0$

Alternative hypothesis: $\mu_D < 0$

Because the difference D is defined as Pretest-Postest. And we want to see if the postets score is higher than the pretest with th training.

The first step is calculate the difference $d_i=x_i-y_i$ and we obtain this:

d: -9, -6, -6, -1, 1, -2

The second step is calculate the mean difference

$\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{-23}{6}=-3.833$

The third step would be calculate the standard deviation for the differences, and we got:

$s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =3.763$

The 4 step is calculate the statistic given by :

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{-3.833 -0}{\frac{3.763}{\sqrt{6}}}=-2.494$

The next step is calculate the degrees of freedom given by:

$df=n-1=6-1=5$

Now we can calculate the p value, since we have a left tailed test the p value is given by:

$p_v =P(t_{(5)}$

The p value is lower than the significance level assumed $\alpha=0.05$ , so then we can conclude that we can reject the null hypothesis. So we can say that the differences between Pretest and Posttest $\mu_{pretest}-\mu_{postest}$ are less than 0.