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3 years ago

P = $14,300 r = 7 % t = 4

1 answer:

5 0

$A=P\left(1+\dfrac{t}{100}\right)^t\\\\P=\$14,300\\r=7\%\\t=4\\\\A=\$14,300\left(1+\dfrac{7}{100}\right)^4=\$14,300(1+0.07)^4\$14,300\cdot1.07^4\approx18,744.38$

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You wish to test whether the number of fouls called in regular season games is different than during the NCAA tournament. The me

kolezko [41]

Answer:

the lower value of the interval = 35.26

Step-by-step explanation:

The mean number is u = 40.1

n = 16

the mean number called x is 38.1

the standard deviation = 5.8

given a 95% Confidence Interval

The lower value of the confidence interval is?

solution

There is a inﬁnite population and the standard deviation of the population is known,

the below formula is used for determining an estimate of the conﬁdence limits of the population mean, i.e.

x ± ( zₐσ)/ √n

For a 95% conﬁdence level, the value of za is taken from the confidence interval table = 1.96.

the conﬁdence limits of the population=

x ± ( zₐσ)/ √n

38.1 ± (1.96*5.8)/ √16

38.1 ± 11.368/4

38.1 ± 2.842

40.942 or 35.258

Thus, the 95% conﬁdence limits are 40.942 or 35.258

this prediction is made with conﬁdence that it will be correct nine five times out of 100.

finally, the lower value of the interval = 35.26

7 0

4 years ago

Can you help me? Please

Rudiy27

6. When one number is a multiple of the other number, that would make the first number the LCM. For example if your first number is 14 and the second number is 2, their first common multiple would be when 2 gets to 14.
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5. The first thing to do to find out if 8 is a multiple of 56 is to divide 56 by 8 and see if it comes out to a whole number. 56/8=7
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6 0

3 years ago

Solve the equation C/4-5=4 Show all work

Anastaziya [24]

C/4-5=4
Add 5 to both sides
C/4=9
Multiply 4 on both sides
Final Answer: C=36

3 0

3 years ago

Read 2 more answers

I simply don’t understand. I need to simplify it

user100 [1]

Answer: i - 60 ${5}^y^{5}$ $z^{8}$ $\sqrt{z}$

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Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

2. In how many ways can 3 different novels, 2 different mathematics books and 5 different chemistry books be arranged on a books

insens350 [35]

The number of ways of the books can be arranged are illustrations of permutations.

When the books are arranged in any order, the number of arrangements is 3628800
When the mathematics book must not be together, the number of arrangements is 2903040
When the novels must be together, and the chemistry books must be together, the number of arrangements is 17280
When the mathematics books must be together, and the novels must not be together, the number of arrangements is 302400

The given parameters are:

$\mathbf{Novels = 3}$

$\mathbf{Mathematics = 2}$

$\mathbf{Chemistry = 5}$

(a) The books in any order

First, we calculate the total number of books

$\mathbf{n = Novels + Mathematics + Chemistry}$

$\mathbf{n = 3 + 2 + 5}$

$\mathbf{n = 10}$

The number of arrangement is n!:

So, we have:

$\mathbf{n! = 10!}$

$\mathbf{n! = 3628800}$

(b) The mathematics book, not together

There are 2 mathematics books.

If the mathematics books, must be together

The number of arrangements is:

$\mathbf{Maths\ together = 2 \times 9!}$

Using the complement rule, we have:

$\mathbf{Maths\ not\ together = Total - Maths\ together}$

This gives

$\mathbf{Maths\ not\ together = 3628800 - 2 \times 9!}$

$\mathbf{Maths\ not\ together = 2903040}$

(c) The novels must be together and the chemistry books, together

We have:

$\mathbf{Novels = 3}$

$\mathbf{Chemistry = 5}$

First, arrange the novels in:

$\mathbf{Novels = 3!\ ways}$

Next, arrange the chemistry books in:

$\mathbf{Chemistry = 5!\ ways}$

Now, the 5 chemistry books will be taken as 1; the novels will also be taken as 1.

Literally, the number of books now is:

$\mathbf{n =Mathematics + 1 + 1}$

$\mathbf{n =2 + 1 + 1}$

$\mathbf{n =4}$

So, the number of arrangements is:

$\mathbf{Arrangements = n! \times 3! \times 5!}$

$\mathbf{Arrangements = 4! \times 3! \times 5!}$

$\mathbf{Arrangements = 17280}$

(d) The mathematics must be together and the chemistry books, not together

We have:

$\mathbf{Mathematics = 2}$

$\mathbf{Novels = 3}$

$\mathbf{Chemistry = 5}$

First, arrange the mathematics in:

$\mathbf{Mathematics = 2!}$

Literally, the number of chemistry and mathematics now is:

$\mathbf{n =Chemistry + 1}$

$\mathbf{n =5 + 1}$

$\mathbf{n =6}$

So, the number of arrangements of these books is:

$\mathbf{Arrangements = n! \times 2!}$

$\mathbf{Arrangements = 6! \times 2!}$

Now, there are 7 spaces between the chemistry and mathematics books.

For the 3 novels not to be together, the number of arrangement is:

$\mathbf{Arrangements = ^7P_3}$

So, the total arrangement is:

$\mathbf{Total = 6! \times 2!\times ^7P_3}$

$\mathbf{Total = 6! \times 2!\times 210}$

$\mathbf{Total = 302400}$