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3 years ago

What is 900 divided by 45

Mathematics

2 answers:

MatroZZZ [7]3 years ago

6 0

I believe the answer is 20

Irina-Kira [14]3 years ago

4 0

Answer:

Step-by-step explanation:

math lol and it's the correct answer

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Negative four and one third ÷ two and one fifth

3241004551 [841]

The solution to the given expression negative four and one third ÷ two and one fifth is -1 32/33

<h3>Fraction division</h3>

negative four and one third ÷ two and one fifth

-4 1/3 ÷ 2 1/5

= -13/3 ÷ 11/5

Multiply by the reciprocal of 11/5

The reciprocal of 11/5 is 5/11

= -13/3 × 5/11

= (-13 × 5) / (3 × 11)

= -65 / 33

= -1 32/33

Therefore, the solution to the fraction is -1 32/33

Learn more about fraction:

brainly.com/question/11562149

#SPJ1

5 0

1 year ago

Read 2 more answers

Let μ denote the true average radioactivity level(picocuries per liter). The value 5 pCi/L is considered thedividing line betwee

VMariaS [17]

Answer:

H0: μ = 5 versus Ha: μ < 5.

Step-by-step explanation:

Given:

μ = true average radioactivity level(picocuries per liter)

5 pCi/L = dividing line between safe and unsafe water

The recommended test here is to test the null hypothesis, H0: μ = 5 against the alternative hypothesis Ha: μ < 5.

A type I error, is an error where the null hypothesis, H0 is rejected when it is true.

We know type I error can be controlled, so safer option which is to test H0: μ = 5 vs Ha: μ < 5 is recommended.

Here, a type I error involves declaring the water is safe when it is not safe. A test which ensures that this error is highly unlikely is desirable because this is a very serious error. We prefer that the most serious error be a type I error because it can be explicitly controlled.

7 0

3 years ago

Two problems here I need solved! I need every step, so please have that with your answers!!

Free_Kalibri [48]

QUESTION 1

We want to solve,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-6x+8}$

We factor the denominator of the fraction on the right hand side to get,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-4x - 2x+8}.$

This implies
$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x(x-4) - 2(x - 4)}.$

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{(x-4)(x - 2)}$

We multiply through by LCM of
$(x-4)(x - 2)$

$(x - 2) + x(x-4) = 2$

We expand to get,

$x - 2 + {x}^{2} - 4x= 2$

We group like terms and equate everything to zero,

${x}^{2} + x - 4x - 2 - 2 = 0$

We split the middle term,

${x}^{2} + - 3x - 4 = 0$

We factor to get,

${x}^{2} + x - 4x- 4 = 0$

$x(x + 1) - 4(x + 1) = 0$

$(x + 1)(x - 4) = 0$

$x + 1 = 0 \: or \: x - 4 = 0$

$x = - 1 \: or \: x = 4$

But
$x = 4$
is not in the domain of the given equation.

It is an extraneous solution.

$\therefore \: x = - 1$
is the only solution.

QUESTION 2

$\sqrt{x+11} -x=-1$

We add x to both sides,

$\sqrt{x+11} =x-1$

We square both sides,

$x + 11 = (x - 1)^{2}$

We expand to get,

$x + 11 = {x}^{2} - 2x + 1$

This implies,

${x}^{2} - 3x - 10 = 0$

We solve this quadratic equation by factorization,

${x}^{2} - 5x + 2x - 10 = 0$

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

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

$x + 2 = 0 \: or \: x - 5 = 0$

$x = - 2 \: or \: x = 5$

But
$x = - 2$
is an extraneous solution

$\therefore \: x = 5$