What sets of numbers do -7/10 belong to

The ratio of the geometric mean and arithmetic mean of two numbers is 3:5, find the ratio of the smaller number to the larger nu

IgorC [24]

Answer:

$\frac{1}{9}$

Step-by-step explanation:

Let the numbers be x,y, where x>y

The geometric mean is

$\sqrt{xy}$

The Arithmetic mean is

$\frac{x + y}{2}$

The ratio of the geometric mean and arithmetic mean of two numbers is 3:5.

$\frac{ \sqrt{xy} }{ \frac{x + y}{2} } = \frac{3}{5}$

We can write the equation;

$\sqrt{xy} = 3$

or

$xy = 9 - - - (2)$

l

and

$\frac{x + y}{2} = 5$

or

$x + y = 10 - - - (2)$

Make y the subject in equation 2

$y = 10 - x - - - (3)$

Put equation 3 in 1

$x(10 - x) = 9$

$10x - {x}^{2} = 9$

${x}^{2} - 10x + 9 = 0$

$(x - 9)(x - 1) = 0$

$x =1 \: or \: 9$

When x=1, y=10-1=9

When x=9, y=10-9=1

Therefore x=9, and y=1

The ratio of the smaller number to the larger number is

$\frac{1}{9}$

3 0

3 years ago

1. There are 78 sophomores at a school. Each is required to take at least one year of either chemistry or physics, but they may

SVETLANKA909090 [29]

We are given that there are a total of 78 students. If we set the following variables:

$\begin{gathered} C=\text{students only in chemestry} \\ P=\text{students only in physics} \\ PC=\text{students in physics and chemistry} \end{gathered}$

Then, the sum of all of these must be 78, that is:

$C+P+PC=78$

Since there are 15 in chemistry and physics and 47 in chemistry, we may replace that into the equation and we get:

$47+P+15=78$

Simplifying:

$62+P=78$

Now we solve for P by subtracting 62 on both sides:

$\begin{gathered} 62-62+P=78-62 \\ P=16 \end{gathered}$

Therefore, there are 16 students in physics

3 0

1 year ago

The n term of a geometric sequence is denoted by Tn and the sum of the first n terms is denoted by Sn.Given T6-T4=5/2 and S5-S3=

Leno4ka [110]

1 step: $S_{5}=T_{1}+T_{2}+T_{3}+T_{4}+T_{5}$ , $S_{3}=T_{1}+T_{2}+T_{3}$ , then
$S_{5}-S_{3}=T_{4}+T_{5}=5$ .

2 step: $T_{n}=T_{1}*q^{n-1}$ , then
$T_{6}=T_{1}*q^{5}$
$T_{5}=T_{1}*q^{4}$
$T_{4}=T_{1}*q^{3}$
$T_{3}=T_{1}*q^{2}$
and $\left \{ {{T_{6}-T_{4}= \frac{5}{2} } \atop {T_{5}+T_{4}=5}} \right.$ will have form $\left \{ {{T_1*q^{5}-T_{1}*q^{3}= \frac{5}{2} } \atop {T_{1}*q^{4}+T_{1}*q^{3}=5} \right.$ .

3 step: Solve this system $\left \{ {{T_1*q^{3}*(q^{2}-1)= \frac{5}{2} } \atop {T_{1}*q^{3}*(q+1)=5} \right.$ and dividing first equation on second we obtain $\frac{q^{2}-1}{q+1}= \frac{ \frac{5}{2} }{5}$ . So, $\frac{(q-1)(q+1)}{q+1} = \frac{1}{2}$ and $q-1= \frac{1}{2}$ , $q= \frac{3}{2}$ - the common ratio.

4 step: Insert $q= \frac{3}{2}$ into equation $T_{1}*q^{3}*(q+1)=5$ and obtain $T_{1}* \frac{27}{8}*( \frac{3}{2}+1 ) =5$ , from where $T_{1}= \frac{16}{27}$ .

5 0

3 years ago

A random sample of 10161016 adults in a certain large country was asked "Do you pretty much think televisions are a necessity o

Korolek [52]

Answer:

0.509

Step-by-step explanation:

Given that a random sample of 1016 adults in a certain large country was asked "Do you pretty much think televisions are a necessity or a luxury you could do without?"

Of the 1016 adults surveyed, 517 indicated that televisions are a luxury they could do without.

To find out the point estimate of population proportion of adults in the country who believe that televisions are a luxury they could do without.

We find that point estimate is nothing but sample proportion

Sample proportion = favourably voters/total people surveyed

= $\frac{517}{1016} \\=0.5088\\=0.509$

6 0

3 years ago

Which of the following equations is true? 3 + 1 = 3 . 0

HACTEHA [7]

You only put one equation in

5 0

3 years ago