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Liula [17]
4 years ago
6

write the following ratio using two other notations 4/9. use only the numbers above (not any others).​

Mathematics
2 answers:
Marysya12 [62]4 years ago
8 0

Answer:

4 : 9 or 8 :18

~Hope this helps!!

NNADVOKAT [17]4 years ago
5 0
4 : 9. and ‘4 to 9’

GOOD LUCK!
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Shiny White dental insurance costs $258 per year. Approximately one-third of insured people need a filling, which averages $110
sveta [45]

Answer:

  about $145.33

Step-by-step explanation:

Consider a group of 15 customers. They will pay ...

  15 × $258 = $3870

in premiums each year.

One-third of those, 5 customers, will submit claims for fillings, so will cost the insurance company ...

  5 × $110 = $550

And 80% of them, 12 customers, will submit claims for preventive check-ups, so will cost the company ...

  12 × $95 = $1140

The net income from these 15 customers will be ...

  $3870 -550 -1140 = $2180

Then the average income per customer is this value divided by the 15 customers in the group:

  $2180/15 = $145.33

_____

<em>Alternate solution</em>

Above, we chose a number of customers that made 1/3 of them and 4/5 of them be whole numbers. You can also work with one premium and the probability of a claim:

  258 - (1/3)·110 - 0.80·95 = 145.33

8 0
3 years ago
Read 2 more answers
Which values of c will cause the quadratic equation -- x2 + 3x + c = 0 to have no real number solutions? Check all that
Novay_Z [31]

Answer:

c=5x

Step-by-step explanation:

:D good luck

6 0
3 years ago
Find the critical points of the function and use the First Derivative Test to determine whether the critical point is a local mi
Andre45 [30]

Answer:

Critical points are 1 and -1

Maximum at x=1

Minimum at x=-1

Step-by-step explanation:

We are given that a function

f(x)=3tan^{-1}(x)-\frac{3}{2} x+5 on (-\infty,\infty)

We have to find the critical points of the function.

To find the critical point we will differentiate function w.r.t x and then substitute f'(x)=0

f'(x)=\frac{3}{1+x^2}-\frac{3}{2}

\frac{d(tan^{-1}x)}{dx}=\frac{1}{1+x^2}

f'(x)=0

\frac{3}{1+x^2}-\frac{3}{2}=0

\frac{3}{1+x^2}=\frac{3}{2}

1+x^2=2

x^2=2-1=1

x=\pm1

Therefore, the critical points of the given function are 1 and -1.

f(0)=3-\frac{3}{2}=\frac{3}{2}

f'(1)=0

f'(2)=\frac{3}{5}-\frac{3}{2}=-\frac{9}{10}

When we goes from 0 to 2 then the sign of derivative  change from positive to negative .Therefore, function has local maximum at x=1.

f(-2)=\frac{3}{5}-\frac{3}{2}=-\frac{9}{10}

f(-1)=0

f(0)=\frac{3}{2}

When we goes form -2 to 0 then the sign of derivative change from negative to positive .Hence , function has local minimum at x=-1

Hence, critical points are local maximum and local minimum .

3 0
3 years ago
What is the answer to this question. Please give me the answer ASAP
strojnjashka [21]

still, if you have any doubt ask me!

6 0
3 years ago
Which of the following examples could correctly show the mean density of lunar soil?
viva [34]
<span>The correct answer is 3.24 grams per cubic centimetre. As a matter of fact, the lunar soil’s mean density may range from 2.90 to 3.24 g cm-3 which highly depends on the type of ingredients present in its resource materials.</span>
7 0
4 years ago
Read 2 more answers
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