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3 years ago

8

Since 2005, the amount of money spent at restaurants in a certain country has increased at a rate of 8% each year, In 2005, abou

t $360 billion was spent at
restaurants. If the trend continues, about how much will be spent at restaurants in 2017?

1 answer:

Kryger [21]3 years ago

3 0

Answer:

This is an exponential growth problem.

The growth rate is 4% (0.04).

The time is 10 years (10 years from 2005 to 2015)

A = P(1+i)t

A = amount spent in 2015

P = amount spent in 2005

i = interest rate expressed as a decimal

t = # years

A = 500(1.04)10

A = 500(1.480244)

A = 740.122 ==> 740 billion

Step-by-step explanation:

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PLEASE HELP ILL MARK BRAINLIEST !!!

Alexus [3.1K]

Answer:

a) the common difference is 20

b) $x_8=115 , x_{12}=195$

c) the common difference is -13

d) $a_{12}=52, a_{15}=13$

Step-by-step explanation:

a) what is the common difference of the sequence xn

Looking at the table, we get x_3=16, x_4=36 and x_5= 56

Deterring the common difference by subtracting x_4 from x_3 we get

36-16 =20

So, the common difference is 20

b) what is x_8? what is x_12

The formula used is: $x_n=x_1+(n-1)d$

We know common difference d= 20, we need to find $x_1$

Using $x_3=16$ we can find $x_1$

$x_n=x_1+(n-1)d\\x_3=x_1+(3-1)d\\15=x_1+2(20)\\15=x_1+40\\x_1=15-40\\x_1=-25$

So, We have $x_1 = -25$

Now finding $x_8$

$x_n=x_1+(n-1)d\\x_8=x_1+(8-1)d\\x_8=-25+7(20)\\x_8=-25+140\\x_8=115$

So, $\mathbf{x_8=115}$

Now finding $x_{12}$

$x_n=x_1+(n-1)d\\x_{12}=x_1+(12-1)d\\x_{12}=-25+11(20)\\x_{12}=-25+220\\x_{12}=195$

So, $\mathbf{x_{12}=195}$

c) what is the common difference of the sequence $a_m$

Looking at the table, we get a_7=104, a_8=91 and a_9= 78

Deterring the common difference by subtracting a_7 from a_8 we get

91-104 =-13

So, the common difference is -13

d) what is a_12? what is a_15?

The formula used is: $a_n=a_1+(n-1)d$

We know common difference d= -13, we need to find $a_1$

Using $a_7=104$ we can find $x_1$

$a_n=a_1+(n-1)d\\a_7=a_1+(7-1)d\\104=a_1+7(-13)\\104=a_1-91\\a_1=104+91\\a_1=195$

So, We have $a_1 = 195$

Now finding $a_{12}$ , put n=12

$a_n=a_1+(n-1)d\\a_{12}=a_1+(12-1)d\\a_{12}=195+11(-13)\\a_{12}=195-143\\a_{12}=52$

So, $\mathbf{a_{12}=52}$

Now finding $a_{15}$ , put n=15

$a_n=a_1+(n-1)d\\a_{15}=a_1+(15-1)d\\a_{15}=195+14(-13)\\a_{15}=195-182\\a_{15}=13$

So, $\mathbf{a_{15}=13}$

5 0

3 years ago

A family consists of a father, a mother and 2 children. What is the probability that all 2 children are boys?

pashok25 [27]

Answer:

¼

Step-by-step explanation:

The probability of having a boy is ½ and that of a girl is ½.

Probability of boy, boy is (pb*pb) and given pb to be ½ then we can prove the point as follows

For these two children

The options are as follows

1 boy(first) and 1 girl

2 boys

2 girls

1 girl( first) and 1 boy

These are four possible options and the option for two boys is 1 out of the four.

The probability of 2 boys is ½*½=¼

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What is subtraction?

Mazyrski [523]

Wen you deduct an amount from something

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kkurt [141]

2.64575131106 I don’t know if you need to round the answer or not

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[(1/(3+x))-(1/3)] / x<br> Limit 0

Effectus [21]

First find a common denominator and combine the fractions in the numerator:

$\displaystyle\lim_{x\to0}\frac{\dfrac1{3+x}-\dfrac13}x=\lim_{x\to0}\frac{\dfrac3{3(3+x)}-\dfrac{3+x}{3(3+x)}}x=\lim_{x\to0}\frac{3-(3+x)}{3x(3+x)}$

Now simplify and cancel out all the terms that you can:

$\displaystyle\lim_{x\to0}\frac{3-3-x}{3x(3+x)}=-\frac13\lim_{x\to0}\frac1{3+x}$

Since the remaining expression is continuous as a function of $x$ , you can directly substitute to end up with

$\displaystyle-\frac13\lim_{x\to0}\frac1{3+x}=-\frac13\times\frac1{3+0}=-\frac19$

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3 years ago