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4 years ago

How do you write 403.608 in word form

2 answers:

8 0

4 is in the hundreds place
3 is in the ones place
6 is in the tenths place
8 is in the thousanths place so
four hundred and three point or dot six zero eight or
four hundred and three and six hundred and eight thousanths

Alexxx [7]4 years ago

3 0

403.608 as word form is:

Four hundred three and six hundred eight thousandths.

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Please help me. I really appreciate it if you answered. Thank you! <3

exis [7]

Answer:

$3100 is invested at 9%

$4900 is invested at 11%

Step-by-step explanation:

Let's take "x" be the amount invested at 9%.

(x + 1800) is invested in another account at 11%.

The interest amount earned by the two accounts is $818.

Here we can use the simple interest formula and find the amount invested in each account.

Simple interest (I) = , where P- is the principal , N is the number of years and R is the interest rate.

Simple interest =

0.09x + 0.11(x+1800) = 818

Now we have to simplify and find the value of x .

Use the distributive property and simplify the second term.

0.09x + 0.11x + 198 = 818

0.2x + 198 = 818

0.2x =818 - 198

0.2x = 620

x = 620/0.2

x = 3100.

So $3100 is invested at 9%

x + 1800 = 3100 + 1800

= $4900

$4900 is invested at 11%

Hope this helped.

6 0

4 years ago

What is the volume of the cylinder below?

3241004551 [841]

Answer:

Step-by-step explanation:

It is C.

Volume-πr^2h

r-11

h-15

11^2*15

=1815π

6 0

3 years ago

Read 2 more answers

student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te

alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = $\frac{1}{4}$ .

Thus, the probability that the student does not receive version A = $1-\frac{1}{4}$ = $\frac{3}{4}$ .

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

$\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}$