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4 years ago

The product of two numbers is 45 and their Arithmetic mean is 7. find the numbers

2 answers:

8 0

Xy = 45
(x+y)/2=7
---------
x+y = 14
Ah. So we can simplify the problem to, 'the product of two numbers is 45 and their sum is 14'. This shouldn't too hard to guess: the numbers are 5 and 9

ikadub [295]4 years ago

5 0

Let the two numbers be x and y.
Now, according to question
x*y=45 (let this be eq1)and (x+y)/2=7(let this be eq2)
x+y=14
x=14-y
Putting this in eq1
(14-y)*y=45
14y-y^2=45
This is a quadratic equation
y^2-14y+45=0
You can solve a quadratic any way you want. Here factorization method is being used.
y(y-9)-5(y-9)=0
(y-5)(y-9)=0
y can be equal to 9 or 5
Using any of the y values we can find x by placing y in eq1 or 2[preferably 2 because addition]
If y=5 then x=9
If y=9 then x=5

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Please help me figure out this problem

tiny-mole [99]

Answer:

$f(g(x))=f\left(-\dfrac{3x+21}{5}\right)=x\\\\g(f(x))=g\left(-\dfrac{5}{3}x-7\right)=x$

Step-by-step explanation:

Fill in the given expressions and simplify.

$f(g(x))=f\left(-\dfrac{3x+21}{5}\right)=-\dfrac{5}{3}\left(-\dfrac{3x+21}{5}\right)-7\\\\=\left(-\dfrac{5}{3}\right)\left(-\dfrac{3x}{5}\right)-\left(\dfrac{5}{3}\right)\left(-\dfrac{21}{5}\right)-7\\\\=\dfrac{15x}{15}+\dfrac{5\cdot 21}{3\cdot 5}-7=x+7-7=x$

Going the other way, we have ...

$g(f(x))=\displaystyle -\frac{3\left(-\frac{5}{3}x-7\right)+21}{5}=-\frac{-5x-21+21}{5}=x$

5 0

3 years ago

Please help thank you

NISA [10]

Answer:

1 - 3 terms (I'm not really sure because you need to show me the options.)
2 - 1/2, 1, 7

Step-by-step explanation:

A term is a single number, variable, or a number multiplied by a variable.
The coefficients are numerical values that are placed behind a variable to multiple them.

3 0

1 year ago

Read 2 more answers

You need a 30% alcohol solution. On hand, you have a 780 mL of a 15% alcohol mixture. You also have 90%

ycow [4]

Answer:

195 mL of the 90% solution

Step-by-step explanation:

Let the number of mL of 90% mixture = x

Hence, our equation is

780mL × 15% + x mL × 90% = (780mL + x) 30%

117 + 0.9x = 234 + 0.3x

Collect like terms

0.9x - 0.3x = 234 - 117

0.6x = 117

x = 117/0.6

x = 195 mL

Therefore, You will need 195 mL of the 90% solution

5 0

3 years ago

the height of an equivalent triangle is 15cm and it's perimeter is 36cm find the area of the triangle

OlgaM077 [116]

Answer:

The area is 72 cm².

Step-by-step explanation:

Since it is equivalent triangle

h = b

so,

b + b + b (since all sides are equal)= 36 cm

3b = 36 cm

or, b = 36/3

so, b = 12 cm

area of triangle = (1/2)×b×h

= (1/2)×12cm×12cm

= 6cm × 12cm

= 72 cm²

4 0

3 years ago

A book claims that more hockey players are born in January through March than in October through December. The following data sh

astra-53 [7]

Answer:

$\chi^2 = \frac{(67-47.5)^2}{47.5}+\frac{(56-47.5)^2}{47.5}+\frac{(30-47.5)^2}{47.5}+\frac{(37-47.5)^2}{47.5}=18.295$

Now we can calculate the degrees of freedom for the statistic given by:

$df=(categories-1)=4-1=3$

And we can calculate the p value given by:

$p_v = P(\chi^2_{3} >18.295)=0.00038$

Since the p value is very low we have enough evidence to reject the null hypothesis and we can conclude that the players' birthdates are not uniformly distributed throughout the year

Step-by-step explanation:

We need to conduct a chi square test in order to check the following hypothesis:

H0: There is no difference of birthdates distributed throughout the year

H1: There is a difference between birthdates distributed throughout the year

The level of significance assumed for this case is $\alpha=0.05$

The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{total}{4}$

And replacing we got:

$E_{1} =\frac{67+56+30+37}{4}=47.5$

And now we can calculate the statistic:

$\chi^2 = \frac{(67-47.5)^2}{47.5}+\frac{(56-47.5)^2}{47.5}+\frac{(30-47.5)^2}{47.5}+\frac{(37-47.5)^2}{47.5}=18.295$

Now we can calculate the degrees of freedom for the statistic given by:

$df=(categories-1)=4-1=3$

And we can calculate the p value given by:

$p_v = P(\chi^2_{3} >18.295)=0.00038$

Since the p value is very low we have enough evidence to reject the null hypothesis and we can conclude that the players' birthdates are not uniformly distributed throughout the year

3 0

4 years ago

The product of two numbers is 45 and their Arithmetic mean is 7. find the numbers​

The product of two numbers is 45 and their Arithmetic mean is 7. find the numbers