All categories

1 year ago

In which year will 67% of babies be born out of wedlock?

1 answer:

3 0

Based on the trend of the increase in children born out of wedlock, if this trend keeps increasing, the year with 67% of babies born out of wedlock will be 2055 .

<h3>What year will 67% of babies be born to unmarried parents?</h3><h3 />

In 1990, the 28% of children were born out of wedlock and this trend was increasing by 0.6% per year.

If the trend continues, the number of years till 67% of children born out of wedlock will be:

= (67% - 28%) / 0.6%

= 65 years

The year will be:

= 1990 + 65

= 2055

The first part of the question is:

According to the National Center for Health Statistics, in 1990, 28% of babies in the United States were born to parents who were not married. Throughout the 1990s, this percentage increased by approximately 0.6 per year.

Find out more on benefits of marriage at brainly.com/question/12132551.

#SPJ1

You might be interested in

F (x) = (x + 5)^3(x - 9)(x + 1)

Lostsunrise [7]

Answer:

F (x) = (x + 5)^3(x - 9)(x + 1) = 5

Step-by-step explanation:

Hope this helps! :)

4 0

2 years ago

A multiplication expression with a product of 108,064 is shown below. What is the missing digit? four thousand nine hundred twel

hoa [83]

Answer:

???????????

Step-by-step explanation:

sorry cant help

6 0

2 years ago

Write the equation -4x^2+9y^2+32x+36y-64=0 in standard form. Please show me each step of the process!

IgorC [24]

Hey there, hope I can help!

$-4x^2+9y^2+32x+36y-64=0$

$\mathrm{Add\:}64\mathrm{\:to\:both\:sides} \ \textgreater \ 9y^2+32x+36y-4x^2=64$

$\mathrm{Factor\:out\:coefficient\:of\:square\:terms} \ \textgreater \ -4\left(x^2-8x\right)+9\left(y^2+4y\right)=64$

$\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}4$
$-\left(x^2-8x\right)+\frac{9}{4}\left(y^2+4y\right)=16$

$\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}9$
$-\frac{1}{9}\left(x^2-8x\right)+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}$

$\mathrm{Convert}\:x\:\mathrm{to\:square\:form}$
$-\frac{1}{9}\left(x^2-8x+16\right)+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)$

$\mathrm{Convert\:to\:square\:form}$
$-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)$

$\mathrm{Convert}\:y\:\mathrm{to\:square\:form}$
$-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y^2+4y+4\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right)$

$\mathrm{Convert\:to\:square\:form}$
$-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y+2\right)^2=\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right)$

$\mathrm{Refine\:}\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right) \ \textgreater \ -\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y+2\right)^2=1$

$Refine\;once\;more\;-\frac{\left(x-4\right)^2}{9}+\frac{\left(y+2\right)^2}{4}=1$

For me I used
$\frac{\left(y-k\right)^2}{a^2}-\frac{\left(x-h\right)^2}{b^2}= 1$
$As\;\mathrm{it\;\:is\:the\:standard\:equation\:for\:an\:up-down\:facing\:hyperbola}$

I know yours is an equation which is why I did not go any further because this is the standard form you are looking for. I would rewrite mine to get my hyperbola standard form. However the one I have provided is the form you need where mine would be.
$\frac{\left(y-\left(-2\right)\right)^2}{2^2}-\frac{\left(x-4\right)^2}{3^2}=1$

Hope this helps!

4 0

3 years ago

I need help ASAP!! 13 points

OleMash [197]

Answer:

The correct answer is H

Step-by-step explanation:

1/4 of a wall in 2/3 of an hour is 40 min

so a wall takes 160 min which is 2 hours and 40 min

so 3 walls takes 8 hours

6 0

3 years ago