MATHEMATICS POST I) Find the interest on $1,200 at 5 1/2% annual interest for 9 months,

Here plz answer and I’m bored ahf

Ann [662]

Answer:

5.5 years aka 5 years 6 months

Step-by-step explanation:

x over 3330

then 2 over 100

2x3000= 6000/100=60

3330-3000=330

330/60=5.5

7 0

3 years ago

Read 2 more answers

Help fast please !!!

postnew [5]

Answer:

i am not trying to get u wrong i honestly dont know srry

Step-by-step explanation:

sorrry :(

4 0

3 years ago

PLEASE HELP. please give it as a proof

Morgarella [4.7K]

AD=DC, because BD is a bisector of AC.

BD is in common.

AD and BD are perpendicular, and so are DC and BD, because BD is perpendicular to AC.

AB^2 = AD^2+BD^2 because ABD is right

BC^2 = DC^2+BD^2 because BDC is right

AB=BC because AD=DC

So, AD=DC, AB=BC and BD is common, and the triangles are congruent.

8 0

4 years ago

An experiment on memory was performed, in which 16 subjects were randomly assigned to one of two groups, called "Sentences" or "

FromTheMoon [43]

Answer:

There is no significant difference between the averages.

Step-by-step explanation:

Let's call

$\large X_{sentences}$ the mean of the “sentences” group

$\large S_{sentences}$ the standard deviation of the “sentences” group

$\large X_{intentional}$ the mean of the “intentional” group

$\large S_{intentional}$ the standard deviation of the “intentional” group

Then, we can calculate by using the computer

$\large X_{sentences}=28.75$

$\large S_{sentences}=3.53553$

$\large X_{intentional}=31.625$

$\large S_{intentional}=1.40788$

$\large X_{sentences}-X_{intentional}=28.75-31.625=-2.875$

The standard error of the difference (of the means) for a sample of size 8 is calculated with the formula

$\large \sqrt{(S_{sentences})^2/8+(S_{intentional})^2/8}$

So, the standard error of the difference is

$\large \sqrt{(3.53553)^2/8+(1.40788)^2/8}=1.34546$

In order to see if there is a significant difference in the averages of the two groups, we compute the interval of confidence of 95% for the difference of the means corresponding to a level of significance of 0.05 (5%).

If this interval contains the zero, we can say there is no significant difference.

Since the sample size is small, we had better use the Student's t-distribution with 7 degrees of freedom (sample size-1), which is an approximation to the normal distribution N(0;1) for small samples.

We get the $\large t_{0.05}$ which is a value of t such that the area under the Student's t distribution outside the interval $\large [-t_{0.05}, +t_{0.05}]$ is less than 0.05.