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3 years ago

2) Using the table above, what is the probability a patient chosen at random will have normal cholesterol level?

Mathematics

1 answer:

prohojiy [21]3 years ago

5 0

Answer:

I cannot answer this. There is no table for me to answer this with. I am sorry I couldn't help you.

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Multiply and simplify: (2b2+2b)2

sveta [45]

Answer:

4b^2+2b (b^2 means b squared)

Step-by-step explanation:

4 0

3 years ago

The price of a home is $300,000. After a 10% down payment, the mortgage is $270,000. If this is financed with a 30-year fixed-ra

irinina [24]

Answer:

The correct solution is "$1618.79".

Step-by-step explanation:

The given values are:

Home price,

= $300,000

Down payment,

= 10%

Mortgage,

= $270,000

Rate,

r = 6%

Time,

t = 30 years

As we know,

⇒ $PV=P[\frac{1-(1+r)^{-n}}{r} ]$

On substituting the values, we get

⇒ $270 000= P[\frac{1-(1+\frac{6}{1200} )^{-12\times 30}}{\frac{6}{1200} } ]$

⇒ $P=1618.79$ ($)

7 0

2 years ago

How do I find the center & radius, then write a standard equation for a,b, and c?

Ghella [55]

Firstly , we will locate centres and radius in the figures

(a)

We can see that

centre=(h,k)=(0,0)

radius=r=5

now, we can use standard equation of circle

$(x-h)^2+(y-k)^2=r^2$

we can plug values

we get

$(x-0)^2+(y-0)^2=5^2$

$x^2+y^2=25$

(b)

We can see that

centre=(h,k)=(5,4)

radius=r=3

now, we can use standard equation of circle

$(x-h)^2+(y-k)^2=r^2$

we can plug values

we get

$(x-5)^2+(y-4)^2=3^2$

$(x-5)^2+(y-4)^2=9$

(c)

We can see that

centre=(h,k)=(4,3)

radius=r=5

now, we can use standard equation of circle

$(x-h)^2+(y-k)^2=r^2$

we can plug values

we get

$(x-4)^2+(y-3)^2=5^2$

$(x-4)^2+(y-3)^2=25$

5 0

3 years ago

What one is it?? Because I am having a hard time figuring it out. If anyone could help it would mean a lot, thanks

hichkok12 [17]

Answer:

the first one

Step-by-step explanation:

5 0

2 years ago

a student is given a reading list of ten books from which he must select two for an outside reading requirement. in how many way

Lelechka [254]

Answer:

The ways in which he can make selection = $10\times 9 =90$

Step-by-step explanation:

Student is having a list of $10$ books.

Assuming that the student can't read the same book twice.

The student has $2$ choices to make for an outside reading requirement,for his 1st choice, he had $10$ books to choose from; for his 2nd, he has $9$ books left to choose from therefore the total permutations of his choices are $=10\times 9 =90$

So the total ways in which the student can make his selection = $90$

6 0

3 years ago